#groups-rings-fields

1 messages · Page 20 of 1

next obsidian
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there’s a few definitions you could be using

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One is that it’s literally defined as Q(sqrt(2))(sqrt(3))

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In which case there’s nothing to prove

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Another is the smallest subfield containint sqrt(2),sqrt(3), and Q

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In which case the proof is near-immediate but it isn’t tautological

hollow parrot
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We do not have the former definition

next obsidian
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But unless you have a definition for it fixed then you can’t prove anything because something is undefined

hollow parrot
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I mean it is pretty clear using the containment of the three

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We just didn't have a definition for multiple elements so I was not sure if that was valid

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Do I need to prove this?

next obsidian
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I mean you can’t prove anything until you have a definition lol

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Like can you prove to me a zuppyshing is always abelian?

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Not until I define what a zuppyshing is

hollow parrot
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My lecture notes make me cry

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We don't use a textbook

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So there's sometimes gaps like this

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Anyway, thank you I am comfortable with this using the second definition

next obsidian
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I mean if you had to pick a definition I guess I’d pick the one being smallest subfield containing

hollow parrot
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Which should be fine

next obsidian
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But the proof is like near immediate

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So idk what was intended

hollow parrot
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I don't necessarily have to prove this the exercise makes use of this fact nonchalantly and I didn't want to just write "clearly" blah blah blah

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But maybe I should

civic tapir
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A somewhat wishy-washy question:

Why is it possible that $(Z_{p^{m+t}} \oplus Z_{p^{n-t}}) / Z_{p^m}$ is cyclic?
I have seen an example, however, it seems like we're recovering structure that didn't exist.
Namely, if $n>m+t$, then $Z_{p^{m+t}} \oplus Z_{p^{n-t}}$ doesn't have an element of order $p^n$ but its quotient does.
I think I might just be having a lapse in intuition

cloud walrusBOT
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justAworm 🥺

553046591812927499.png
next obsidian
civic tapir
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that's good to hear

next obsidian
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The latter part

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The order of the image of sometbing always divides the order of the thing mapping to it

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So a quotient always has elements of lesser order

civic tapir
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Awesome, I figured as much

latent island
ruby sundial
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zuppyshings

latent island
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can some1 explain to me this example

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im still q new to this so im kinda lost

ruby sundial
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ikay

latent island
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i think S3 is just all the permutations of mappings from {1,2,3} to {1,2,3}?

ruby sundial
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recite what makes a subgroup normal

latent island
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n it can be represented as a cycle

ruby sundial
latent island
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um the cosets form a group

ruby sundial
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ok finish sentence

latent island
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wait theres more?

ruby sundial
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A subgroup N, of a group G, is normal if (what goes here)?

knotty frigate
ruby sundial
knotty frigate
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wait

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what

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oh

latent island
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omg

latent island
ruby sundial
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We have a couple of definitions

knotty frigate
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you mean can be represented using cycle notation?

latent island
knotty frigate
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ah ok

latent island
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sry is cycle and cycle notation diff

ruby sundial
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gN=Ng or gNG^-1=N or similar things

latent island
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idk i havent heard of cycle so i just called it that

latent island
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that equivalent i think? im still a bit fuzzy on how

knotty frigate
latent island
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ok anyway

ruby sundial
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So if you take (1),(123),(132) and multiply by the right by any element g and the left by g^-1 it should be one of those three above

latent island
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umm

ruby sundial
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some element g is an element of S_3.

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ill give you an example

latent island
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yes please

ruby sundial
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firstly do you know how to compute cycles

latent island
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um

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u go from the number 1

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or the first element

ruby sundial
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my bad

latent island
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follow where it maps too recursively

ruby sundial
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compute multiplication of cycles

latent island
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i think we go from the right most cycle

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in each cycle we see if our number shows up

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if it does we take the next right number (or first if its the rightmost)

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wait sry thats computing the output of the cycles

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i think i saw about multiplication of cycles but no i dont rmb it rn

ruby sundial
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okay ill give an example

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and ill break it down if u wanna

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in S_6

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(1 3 5)(2 4 3)

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what does this equal

latent island
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wait

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i just googled it

ruby sundial
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ok

latent island
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i think ik lemme try

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(1 3 5) (2 4

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wait sry

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smt died

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halp plz

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(1 2 4 5)

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AH THIS?

knotty frigate
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close

latent island
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D:

knotty frigate
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what happens to 1 as it goes through the first permutation

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(2 4 3)

latent island
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1 becomes 3

ruby sundial
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yeah

knotty frigate
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well you skipped my question but whatever

ruby sundial
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just track where 1 goes

knotty frigate
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yeah

latent island
ruby sundial
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nah lemme show u

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then u get 1 more

knotty frigate
ruby sundial
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look at 1

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we read from right to left

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so

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1 in (243) goes nowhere

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1 in (135) goes to 3

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we have “(13 “so far

latent island
ruby sundial
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3 goes to 2 in (243)

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2 goes nowhere in (135)

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So we have (132

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2 goes to 4 in (243) and nowhere in (135)

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so we have (1324

latent island
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sry my brain pain

ruby sundial
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i can draw it

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but where are u lost so far

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its an algorithmic way of multiplying

latent island
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ok i think i see it now

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i just messed up a lot of times earlier

ruby sundial
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okay

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fine

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now

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new example

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S_8

latent island
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yes

ruby sundial
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(1 2 5 6)(1 3 4 7 8)

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and another one in S3

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( 1 2)(1 2 3)

latent island
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(1 3 4 7 8)(2 5 6)

latent island
ruby sundial
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yes

knotty frigate
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oh wow

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cool

ruby sundial
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keep K as the fixed field

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wait intuition is telling me yes

knotty frigate
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oh

ruby sundial
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yes here is why?

next obsidian
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You can do it for finite extensions easily then glue those because finite extensions of K exhaust the entire field

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I think

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Induction

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Or do something writing a transcendence basis, you should be able to do it for algebraic extensions pretty simply too

ruby sundial
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yeah nvm it would depend on what type of extension for K/F and K/F’

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atlesst what i was doing

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i was assuming seperable i think

next obsidian
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Look it up in Dummit and Foote or something

knotty frigate
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lets say f:F->F' was a isomorphism
can we make a g:K->K such that g(x)=f(x) for all x in F and such that g is an isomorphism

next obsidian
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You write down a set of generators

knotty frigate
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same q right?

next obsidian
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Say there’s n

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You first extend up when adjoining only one of them

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This is simple to do

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Then you do the other n-1 via induction

latent island
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what happened to what i was doing

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welp

ruby sundial
latent island
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both worked?

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i dont like this

next obsidian
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Just look it up in D&F

latent island
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D:

ruby sundial
latent island
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can we move to a help channel

next obsidian
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It’s somewheee in the field theory section

ruby sundial
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nah

ruby sundial
latent island
ruby sundial
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what happens if you multiply g(123)g^-1

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let g be (1 2)

latent island
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wait sry

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whats

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g^-1

ruby sundial
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inverse of g

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for g an element of the group

latent island
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is it just (1 2)

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in this case

ruby sundial
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yeah

latent island
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then if g is

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(1 2 3)

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g^-1 is

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(1 3 2)?

ruby sundial
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yes

latent island
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g(123)g^-1

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(12)(123)(12)

ruby sundial
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yes

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compute that

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it should be one of the elements in your group

latent island
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(13)

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this?

ruby sundial
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i think you are missing something

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the product isnt (1 3) on my end

latent island
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hm?

ruby sundial
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remember we computed (12)(123) before?

latent island
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(12)(123)(12)

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when 1 goes through

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we get like

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(12)(123)(12) 1

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(12)(123)2(12)

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(12)3(123)(12)

ruby sundial
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yes

latent island
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3(12)(123)(12)

ruby sundial
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yup

latent island
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is this a proper notation or just weird lol

ruby sundial
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it helps you keep track

latent island
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anyway 3 goes back to 1 and i think thats it?

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wait a sec no

ruby sundial
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no

latent island
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(12)(123)(12)3

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(12)1(123)(12)

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2(12)(123)(12)

ruby sundial
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?

latent island
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(132)?

ruby sundial
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yeah

latent island
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rookie mistake

ruby sundial
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its in our normal group right?

latent island
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wait sry what was the original thing again

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im confused lol

ruby sundial
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what was your original question

latent island
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tryna understand this

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from here

ruby sundial
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yes

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look at N

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now we just computed an example

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but essentially ghg^-1 is in N for every g in S_3 and h in N

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thats what it means to be normal

latent island
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hm for

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N = {(1), (123), (132)}

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?

ruby sundial
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thats what it says

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im only reading what you are

latent island
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hm lemme process a bit

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ah this?

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looks a bit different tho

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ghg^-1

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wait no urs is different

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hm

ruby sundial
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abuh

latent island
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nani

ruby sundial
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mine is the second bulletin

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however the commutator one is also true

latent island
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ok sry

ruby sundial
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is n^-1g^-1ng is in N

latent island
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we will worry about that commutator one another time

ruby sundial
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its because g^-1ng is in N

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and n^-1 in N

latent island
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The image of conjugation of {\displaystyle N}N by any element of {\displaystyle G}G is equal to {\displaystyle N.}{\displaystyle N.}

ruby sundial
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N is a subgroup so multiplying the two is in N

latent island
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sry what is this

ruby sundial
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you know

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you dont need to post what it is

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we both see the same thing

latent island
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ok mb

ruby sundial
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the second bulletin in the image you posted

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gng^-1 is called conjugation of n by g.

latent island
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OH WAIT WHAT

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ok lol

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ok um

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now what tho

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i dont quite get what we showed with the

ruby sundial
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have you heard conjugation used in english?

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because i dont think i have tbh

latent island
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hm nope

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gng^-1 =n

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so with that we showed n is a normal subgroup?

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whats g about

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OH wait n is a normal subgroup while n is a subgroup of g

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ok wait'

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wait where did

ruby sundial
latent island
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(12) come from tho

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isnt g like

ruby sundial
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gng^-1 is in N

latent island
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1, 2, 3

ruby sundial
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so it can be another elemetn in N not n.

latent island
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wait no

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G = {(1), (12), (13), (23), (123), (132)}

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like this?

ruby sundial
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this is S3

latent island
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so we wanna show for any g in G

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gNg^-1 = N

latent island
ruby sundial
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yes

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for every g in G

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saame thing

latent island
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oh

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so alternatively

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a normal subgroup in this case is

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N = {(1)}

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or

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N = G

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so those are the trivial subgroups

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ok cool

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i think i get it now

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thanks

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💕

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Breakthrough

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Level up Spirit Immortal Layer 5

ruby sundial
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how old are you

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if you dont mind me asking

latent island
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wow

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i dont mind answering in dms

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not here

chilly ocean
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Big 🤨 energy from this chat right now.

ruby sundial
rustic crown
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you wanna prove that it's possible to recover K' from the minimial polynomial which you call m_{x, K'}

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in other words, the map {intermediate fields} --> {monic divisors of m_{x, K}} is injective

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as the right side is finite, this would force the left to be finite

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how to prove "you can do it", or how to prove it

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for the first one, you can try proving it, if the proof works out, then it means you just proved that you could do it :3

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for the second, try relating K' with the coefficients of m_{x, K'}

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(do the coefficients generate K'?)

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well the only information contained in the polynomial m_{x, K'} is its coefficients

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and one thing we know for sure is that this is a poynomial in K'[T], so at least all the coefficients are inside K'.

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if you wish to recover K', from the polynomial, you gotta try something with the coefficients, as that's all you have

rustic crown
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they can be, but in this case you then would want to show that K' = K

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if you can show K'' = K', then you're done!

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ig you should try an example then

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take K = Q and x = sqrt(i)

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so x^4 + 1 = 0

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this has intermediate fields Q(sqrt(2))

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and Q(i), and Q(sqrt(-2))

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what are the minimal polynomials of x over each?

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yep!!

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this is the minimal polynomial for i, not x = sqrt(i)

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hehe :p

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and that's pretty much the proof lol

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yea, but the thing here is that K(x) is generated by a single element. this is all that makes the difference

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if F/K is finite and generated by even 2 elements, then you can't show that there are finitely many intermediate fields

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the standard counterexample is K = F_p(s^p, t^p) and F = K(s, t)

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here the fields generated by s + ct will all have degree p. for any c in K

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read this again :3

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what is m_{x, K''}?

chilly ocean
rustic crown
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so the minimal polynomial is has degree at most p

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but it has to divide the degree of the extension [F:K] = p^2

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why do you say so?

rustic crown
# chilly ocean why does this example work?

and i should say that no two c will give you the same intermediate field. if K(s + at) = K(s + bt) = K', say s + at and s + bt are both in K', which means (a-b)t is in K', which means t is also there and so s is also there, but then K' = F is a contradiction

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and that m_{x, K'} lies in K''[T]

rustic crown
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so you get that [K(x) : K'] = [K(x) : K'']

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can you finish the argument?

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yea, my bad... i was answering two things simultaneously >.<

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yee

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lol

pale jasper
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Given y even, how can I show that y+i and y-i are coprime?

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And also get the bezout identity

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i.e. a(y+i) + b(y-i) = 1

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(In Z[i])

rustic crown
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if d divides both, then it divides their difference 2i

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but as y is even, you get d also divides y

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so d must divide i = (y+i) - y which is a unit, so d is a unit

pale jasper
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What about getting the bezout identity?

rustic crown
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just do the same computations as a euclidean algo

pale jasper
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I'm struggling to do that in Z[i]

rustic crown
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(y - i) = (y + i) - 2i

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(y + i) = 2i(-ix) + i

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where y = 2x

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plug the value of 2i from the first equation

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(y+i) = ((y+i) - (y-i)) * (-ix) + i

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(1+ix) * (y+i) - (ix) * (y-i) = i

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(x - i)(y+i) - x(y-i) = 1

pale jasper
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ty

pale jasper
rustic crown
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1+2ix-iy

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and y = 2x

unique valve
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How would you find the splitting field of x^35-1 over F2

mighty spade
unique valve
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Doesn’t that product have degree 12

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Or 10

mighty spade
unique valve
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x^4+x^3+x^2+x+1 is the 5th cylcotomic polynomial right?

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Isn’t x^35-1 the product of the 1st, 5th, 7th, and 35th cyclotomic polynomials

mighty spade
mighty spade
unique valve
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How does this help

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I assume these are all irreducible in F2 but how would you prove that

rustic crown
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the splitting field is an extension of F2, say it's degree is n.

unique valve
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Oh I see what you mean

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So n is at least 12?

rustic crown
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okie maybe it is better to say it like this

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the set of roots of x^35 -1 forms a subgroup

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of order 35 inside F_{2^n}*

unique valve
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How do you know there aren’t repeating roots?

rustic crown
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by the derivative thing

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its derivative is 35x^34

unique valve
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Yes ok

rustic crown
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by lagrange 35 divides 2^n - 1

unique valve
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I am aware of this although I don’t totally understand why this applies to general fields

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Yes

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And the order of 2 is 12 right?

rustic crown
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idk, need to check :p

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2^4 = 1 mod 5

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2^3 = 1 mod 7

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right so if 2^n = 1 mod 35 then 3 and 4 both divide n

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so 12 divides n

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we'll now show that F_{2^12} is the splitting field

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and use that group of units of this is cyclic

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and pick a generator g

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say 2^12 - 1 = 35 * m

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g^{35 m} = 1

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this means h = g^m is a root of the polynomial x^35 - 1

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and order h is 35

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so the roots are exactly {1, h, h^2, ..., h^34}

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which means it does split completely

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so done eeveeKawaii

unique valve
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Ok that makes sense

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Thanks

rustic crown
unique valve
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I’ll need to convince myself later that the derivative thing applies to arbitrary fields

rustic crown
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in general, the same argument shows that the splitting field of x^r - 1 over F_p is F_{p^n} where n is the order of p mod r

rustic crown
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f doesn't have repeated roots (in its splitting field) if and only if gcd(f, f') = 1

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because if you have a repeated root, then it's a common root of f and f'

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f = (x-a)^2 * g
then f' is divisible by x-a

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and converse is also true

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writing it in terms of gcd is nice because it shows that separability can be checked inside your base field itself and doesn't require you to look at any splitting fields

unique valve
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Yeah bc you don’t need to factor

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That is nice

rustic crown
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but in our case f' = 35 x^34, which only has a single root 0

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and it's not a root of f

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so we could just have did that directly

rustic crown
unique valve
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But isn’t that always dependable

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Seperable

rustic crown
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ah yes, here i wanted r to be coprime to p

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because if r = s * p^k, then x^r - 1 = (x^s - 1)^{p^k}

unique valve
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Ah because then you get 0

rustic crown
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yea :p

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but it's not a huge deal because the splitting field of that is same as x^s-1

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and s is indeed coprime to p

unique valve
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Indeed

rustic crown
unique valve
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Thanks

rotund aurora
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Is the correct way of thinking about universal properties as characterizations of objects (in some context)? For example, a basis for an F-vector space is characterized by the usual universal property.

next obsidian
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Yes

formal ermine
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a couple of questions regarding notation: \
let $A, B$ be submodules of an $R$-module $M$. is $A + B = \Set{a + b | a \in A, b \in B }$? \
what does $M \oplus M'$ mean? \
we defined the direct sum like this: $$\bigoplus_{i \in I} M_i = \Set{(m_i)_{i \in I) | m_i \in M_i, m_i \text{ except for finitely many } i \in I}$$ I don't understand the "except for finitely many" part

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@cloud walrus

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oh

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texit is dead again

formal ermine
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"außer für endlich viele" means "except for finitely many" fwiw

rustic crown
#

what language is that btw catThink

formal ermine
#

german

rustic crown
#

F

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i'm in germany and i still no nothing

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but it did look familiar 😂

formal ermine
#

oh what part of germany?

rustic crown
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muenster

formal ermine
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ah bavaria

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I'm in north rhine westphalia

rustic crown
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(>.< my geography isn't that good, but i think that's the state containing münster, right?)

formal ermine
#

münster is in bayern (bavaria)

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düsseldorf, bochum, dortmund, etc are in nrw

rustic crown
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oh okie, i'll believe you 🙈

formal ermine
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wait what the fuck

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OH

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münchen is in bayern

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jesus christ

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my geography is worse than yours

rotund aurora
formal ermine
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I live an hour away from münster

rustic crown
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lol i my uni name has westphalia in it, so i assumed they were sort of connected 😂

formal ermine
rustic crown
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ah yes :p

rotund aurora
#

are grad schools in germany english based?

rustic crown
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idk, i think they started this is like very recently.. this might be the second year of the program

rustic crown
formal ermine
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ahhhhh

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yeh lol

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it was this

image.png
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but it's the same question

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what does the except for finitely many mean

rustic crown
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right

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it means only finitely many m_i are allowed to be non-zero

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so m_i = 0 is true for almost all i :p

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I'll call M as the direct sum

formal ermine
rustic crown
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so notice that there are natural maps M_i --> M

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and we want any element in M to be sum of finitely many things in images of these maps

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as sum of infinitely many things makes no sense without topology

formal ermine
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wait I have a question before that

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I don't understand the notation $(M_i){i \in I}$ and $(m_i){i \in I}$

rustic crown
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ah that's a short hand notation for a tuple

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like you write (a_1, a_2) for a pair

formal ermine
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ah ok

rustic crown
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another way to think of this is as a set function from a : {1, 2} --> A

formal ermine
#

ye

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what's a "family of submodules"

rustic crown
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similarly (m_i)_{i in I} is just a function m : I --> union of M_i (where ith thing lands in M_i)

rustic crown
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for a collection of elements people just say a set

formal ermine
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why are they writing "Let (M_i)_i in I be a family of submodules" and not "Let { M_i | i in I } be a set of submodules"

rustic crown
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ofc you can use the same word for collections of sets, it just becomes harder to read

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so people prefer to call it family or class of sets

formal ermine
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ah

rustic crown
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but they just mean a set of sets

formal ermine
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so family = set of sets

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ye

rustic crown
#

ig a family of elements is a set :p

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family of submodules = a set of submodules

formal ermine
#

why are they using the tuple notation here

rustic crown
#

tuples can have repeats

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a set can't

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so direct sum of (M, M) is M⊕M

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ig

formal ermine
#

what's the M oplus M notation

rustic crown
#

but it doesn't matter much, as people know what you mean when you write {M_i}_{i in I}

rustic crown
#

M⊕N = {(m, n) : m in M, n in N} = M x N

formal ermine
#

ahhhhhh M oplus N is just a fancy way of saying the cartesian product of M and N?

rustic crown
#

ig

#

but it means more stuff

#

okie lets go from start

#

forget whatever we said till now

#

(also do you know some category theory?)

formal ermine
rustic crown
#

hmm

#

okie so lets think it like this

#

say you have 2 R-modules M and N

#

and you wanna define a product for them

formal ermine
#

yes

rustic crown
#

what do you do?

formal ermine
#

wdym product

rustic crown
#

exactly, what do you mean by that

chilly ocean
elder wave
chilly ocean
#

Funny coincidence in nomenclature

elder wave
#

ain't no way

rustic crown
formal ermine
formal ermine
rustic crown
#

this usually works for all usual algebraic objects you study at the start, because they're just sets with some extra structure

elder wave
#

hm it would probably help to know some linalg

#

the direct sum is a lot more intuitive there imo

rustic crown
#

but there is a dual notion of a product, called a coproduct

rustic crown
#

for example coproduct of sets is their disjoint union

#

(finite) coproduct of abelian groups or R-modules agrees with the product

#

but this notion is a little tricky because coproduct of groups can be weird

#

it's called the free product

formal ermine
#

dual notion?

rustic crown
#

yea

chilly ocean
#

Reverse the arrows

rustic crown
#

just from the example, it might not directly look like product and coproduct are related at all

#

but once you try to give a nice general definition of both, they are just duals of each other

chilly ocean
#

Products and coproducts always come with some maps

formal ermine
#

what is a coproduct

chilly ocean
#

And they are dual because if you reverse the arrows then you get one from the other

rustic crown
#

first let's define a product

formal ermine
#

what does it mean to be a "dual of something"

rustic crown
#

let's work in the case of groups for now

#

the product of G and H is an object G x H such that you have group homomorphism G x H --> G and G x H --> H

chilly ocean
rustic crown
#

and G x H is universal in this sense

formal ermine
#

universal?

rustic crown
#

a universal object mapping to G and to H

#

which means if you have a third group K which also maps to G and to H

chilly ocean
#

This might get a little confusing

rustic crown
#

then you can uniquely factor it through G x H

#

oh wait texit is dead

#

hmm

#

how do i draw diagrams catThink

chilly ocean
#

Using tikz

#

You can use quiver and screenshot

rustic crown
#

lol that's a better idea

#

i was gonna compile locally

#

@formal ermine can you read this diagram?

formal ermine
rustic crown
#

you don't need to understand anything before that

formal ermine
#

what is a universal object mapping

rustic crown
#

think it like an object satisfying certain properties, but only those

#

basically doing the required thing in the most efficient way

formal ermine
#

'an object'?

rustic crown
#

if you have two groups G and H, what does it mean to fine a universal group which has maps to both G and H?

rustic crown
#

i shouldn't use any weird words which might confuse you even more

#

give me one last chance :p

#

Say G and H are groups. We want to define what G x H is completely using groups homomorphisms and not talking about elements at all

#

so one nice thing you want from G x H is to have maps to both G and H

#

(the projection maps)

chilly ocean
formal ermine
formal ermine
chilly ocean
#

but describing this category would be troublesome

formal ermine
#

haven't learned it yet

rustic crown
#

((a, b) -->a and (a, b) --> b)

chilly ocean
#

so it's better to say, it satisfies the following property called the "universal" property

formal ermine
chilly ocean
#

all the things det is talking about here can be perfectly formalized in the language of category theory

rustic crown
#

yep, and if you try to understand it now, it might give you reasons why one likes to think categorically

#

so the first question is, does having maps to G and H determine G x H (upto isomorphism?)

#

and the answer is clearly no

chilly ocean
#

Okay maybe. Imagine you have some sort of structures and maps between them.
You want to figure out how some basic objects in them, like products and coproducts look like.
This is what category theory helps you with

rustic crown
#

because there are sooo many groups that have maps to both G and H

chilly ocean
#

For example, you might want to see what product/coproduct of uniform spaces is. And this can indeed be done

rustic crown
#

for example, for the trivial group {e} also you have group homomorphism {e} --> G and {e} --> H
and certainly you don't wanna say {e} is the product of G and H (if at least one of them is non-trivial)

#

so what extra conditions should one require such that we do characterize G x H uniquely (up to isomorphim)

#

and one answer to this is the following

#

this diagram says that G x H has maps to G and H

#

but it does so in the most natural way

formal ermine
#

what is K

rustic crown
#

which means that if K is any other group also mapping to G and H

#

basically trying to pretend to be a product

chilly ocean
#

So you might think that in the categorical definition of product there is some arbitrary choices here.
But the isomorphism is in fact pretty rigid, so no need to worry about that

rustic crown
#

then there must be a unique group homomorphism K --> G x H such that the diagram commutes

#

does that sort of make sense?

formal ermine
#

I understand the diagram but it makes no intuitive sense to me

rustic crown
#

okie so let's see it in action

#

let's show that the usual way to define product of groups satisfies this property

chilly ocean
#

well - basically we're taking the product and seeing it in terms of properties of what it means to be a product in terms of admissible maps

rotund aurora
#

I was just trying to learn about universal stuff and here det is giving a master class, perfect timing glassescat

next obsidian
rustic crown
#

if you have maps g :K --> G and h : K --> H, can you come up with a map f :K --> G x H such that K --> G x H --> H is g and K --> G x H --> H is h?

chilly ocean
#

like how a map from K to G x H splits into two maps K to G and K to H

next obsidian
#

Like you write a function f:R -> R^n usually like (f_1,…,f_n)

#

This is an example of this, you’re making an object for which maps into it are just maps into the constituent parts

rustic crown
#

(lol 3 people trying to explain the same thing at the same time in 3 different ways 🙈)

#

meanwhile @formal ermine is confused :p

elder wave
#

didn't this all start with just the direct sum

next obsidian
#

I just wanted to give an example

rustic crown
#

yep 😂

elder wave
#

i'm not too sure if this is the way

rustic crown
#

cat pilling everyone is nice

chilly ocean
#

it's educational

elder wave
#

indeed, however will it solve the problem at hand

rustic crown
#

ig :p

formal ermine
#

@chilly ocean I know you're just trying to help, and I am very thankful for that, but you always shoot pigeons with canons

#

like I said I know no category theory and you started talking about category theory :p

rustic crown
#

it will end with why the coproduct in R-modules is that weird thing and why we require only finitely many to be non-zero

chilly ocean
#

everyone started talking about category theory, not just me

rustic crown
#

as opposed to product in R-modules where we don't put the extra condition of almost all being non-zero

rustic crown
rotund aurora
#

What det is saying can be understood without any category theory. But it will suggest the importance of arrows

rustic crown
rustic crown
#

and you can see that it does commute

#

in fact this definition is forced on you

#

because we want the composition K --> G x H --> G to be g

#

so this tells you that if f(k) = (a, b) then a must be g(k)

delicate orchid
rustic crown
#

similarly the other composition tells you b = h(k)

#

(does that make sense? >.<)

#

(if not we'll try again later eeveeKawaii)

#

cat pilling is important uwucat

unique valve
#

Math

#

What’s everyone’s favorite finite group?

rustic crown
#

maybe Q_8?

#

its smol and surprising

rotund aurora
#

Z/2Z

south patrol
#

the trivial group 1, I call it the dust group because
"All go unto one place; all are of the dust, and all turn to dust again."*

#

(jk i am not that pretentious)

delicate orchid
#

PSL_2(7)

#

or GL_3(2)

rustic crown
#

dust maps to everything

south patrol
#

and from everything

#

indeed

chilly ocean
#

I think infinite groups are better because you can give them a non-trivial Hausdorff topology

unique valve
#

They are not

rustic crown
#

whuts your favorite topological group

chilly ocean
#

Solenoid

delicate orchid
rotund aurora
#

General linear groups over finite field are pretty cool

delicate orchid
#

discrete topology see if I care

south patrol
rustic crown
chilly ocean
#

it's an inverse limit of circles

#

with the bonding maps being taking some power > 1

#

it's a whole family of topological groups

#

it can be realized in R^3 as well

rustic crown
chilly ocean
#

but rings are interesting too. Questions like, when is C(K) homeomorphic to C(L) for compact Hausdorff spaces K, L, for instance

#

well, for me it'd be more about topological vector spaces

chilly ocean
#

you need to prove that any K-hom is surjective, so you write L as a union of roots of polynomials with coefficients in K (L\K is algebraic) and you know that any K-hom acts transitively on the roots of any fixed polynomial since that's a finite set and K-homs are injective

rotund aurora
#

In contexts like category theory, do you use the epsilon relation $\in$ to express membership to a proper class (just ignoring the problems it might cause)? If not, what expressions are used to express that an object "belongs" to a proper class ?

tender bough
#

Let $\mathfrak{sl}(n)$ be the special linear Lie algebra, it has dimension $n^2 - 1$, the adjoint action $\mathrm{ad}_H$ of some $H \in \mathfrak{sl}(n)$ has codomain $\mathfrak{gl}(\mathfrak{sl}(n))$, what's the dimension of $\mathfrak{gl}(\mathfrak{sl}(n))$?

tender bough
#

when can I have my tex bot back?

south patrol
#

i guess this is also nicer even in the finite case right? cause no appeal to dimensionality at all

#

(i was gonna say you can just reduce the problem to the finite case but it's just a longer way of doing what you did lol)

#

By "acts transitively on the roots of any fixed polynomial", don't you mean that it sets roots to roots in a bijective fashion though? Maybe it's using the word "transitively" in a different sense to what I'm used to though, but here this is just one fixed map rather than a group of maps

chilly ocean
south patrol
#

Ah sure, no worries, just checking

#

i didn't even realise joe's statement was true until he pointed it out lol

chilly ocean
#

i think it's something that should be taught in any introductory galois theory course

glossy crag
#

Let a be in L, you want to show a has a preimage under your K-hom f. Collect all the roots of the minimal polynomial of a over K in L, say {a_1,...,a_m}. Then K(a_1,...,a_m)/K is a finite extension and its mapped by f into itself, because f sends every a_i to some a_j (this is the key observation here, it's trivial if you think about it for a second). This is a finite-dimensional vector space and f is injective (as a field homomorphism of L into L), so it's an automorphism of K(a_1,...,a_m)/K and a has a pre-image.

#

Does anyone have a clue why if m is even p should be 2?

image.png
south patrol
#

p^e = m deg g

glossy crag
#

Oh fuck, that's so dumb of me.

#

nvm

south patrol
#

dw

glossy crag
south patrol
#

aha nws it happens

#

we can both delete if you wish anyway lol

glossy crag
#

Nah, it's fine, maybe someone will find it useful.

#

I see you a lot on this channel, you a student?

next obsidian
#

He’s a spud

glossy crag
#

A what?

next obsidian
#

A spud

#

Like, a potato

#

🥔

glossy crag
#

Never heard that word before.

south patrol
glossy crag
#

What level/country?

south patrol
#

3rd year undergrad, UK

#

hbu

glossy crag
south patrol
#

Noice, pleased to meet you

next obsidian
#

Wait potato you’re a 3rd year?

#

Lmfao

#

I thought u were a second

south patrol
#

rip

glossy crag
next obsidian
#

You’re less cracked than I thought

south patrol
#

:(

next obsidian
#

🚬

#

Just kidding :)

south patrol
#

I mean I've only been doing math properly for 2 years

glossy crag
next obsidian
#

Pffft I bet you did further maths

south patrol
glossy crag
#

Don't tell me.

south patrol
#

but eh

glossy crag
#

You were a filthy code monkey before weren't you.

south patrol
#

I wasn't actually

glossy crag
#

Physics?

south patrol
#

But that leaves one option

#

yes

glossy crag
#

It's always one or the other.

#

Except with big-brain Ed "The Phenotype" Witten

south patrol
#

lol

#

history

glossy crag
#

Who was a fucking journalism major or something.

glossy crag
south patrol
#

(i think)

next obsidian
#

It’s very funny for me to imagine CS people at your uni Potato

#

It feels so out of place like

south patrol
#

yeah i mean it is meant to be good at CS but it does feel out of place lol

next obsidian
#

Shouldn’t you only be teaching things that have existed for like 800 years or something

glossy crag
#

Witten attended the Park School of Baltimore (class of '68), and received his Bachelor of Arts degree with a major in history and minor in linguistics from Brandeis University in 1971.[10]

He had aspirations in journalism and politics and published articles in both The New Republic and The Nation in the late 1960s.[11][12] In 1972 he worked for six months on George McGovern's presidential campaign.[13]

witten.jpg
south patrol
#

lmao

tender bough
#

I really don't understand what this technically means. Specifically, I have this question: Say H is in the Cartan subalgebra, and v is an eigenvector of ad_H, then is v also a simultaneous eigenvector? i.e. is v also an eigenvector of ad_H' for all H' in the Cartan subalgebra?

IMG_0366.png
glossy crag
next obsidian
#

Not for me to say

glossy crag
#

I'll show you mine if you show me yours :^)

glossy crag
next obsidian
#

Lol?

#

What does that even mean

glossy crag
#

Who is your daddy and what does he do

#

(what year)

elder wave
#

feds in abstract algebra

next obsidian
#

He buys domestic building products and exports it to Japan

tender bough
glossy crag
glossy crag
#

What's everyone up to, algebraically-speaking? I'm browsing some chapters in Isaacs' Algebra and planning on reading Neukirch's ANT.

south patrol
#

uh i am finishing up and reviewing some galois and rep theory shenanigans

glossy crag
tender bough
south patrol
#

Nice

next obsidian
#

I am reading “The Cotangent Complex of a Morphism”

glossy crag
glossy crag
next obsidian
#

It’s quite something. It also has an incredibly funny typo on like the first page

#

The names of the maps are supposed to be shifted back one lol

glossy crag
glossy crag
unique valve
#

Level 2

next obsidian
#

Idk

#

It was cutting edge in the 60s

glossy crag
#

No, I mean you an undergrad/grad or smth else

next obsidian
#

1st year graduate

glossy crag
#

Nice. I was speaking to a professor here the other day and apparently in Europe (or at the very least Germany) graduates (aka PhD students) aren't expected to take classes, just do research.

next obsidian
#

Because in Europe you do a master’s first

glossy crag
#

True, I didn't think about it before, but you guys don't have one.

#

For you it's both in one.

next obsidian
#

For the most part yes

agile burrow
#

but that's just me going off linear algebra, idk much about lie algebras

glossy crag
#

Just possibly scale them differently?

agile burrow
#

oh maybe you're right

#

i forgor about that

glossy crag
#

I don't remember what the requirement for SD was, wasn't it just there's a fixed basis and it's an eigenbasis for each operator?

agile burrow
#

each diagonalizable and commute

glossy crag
#

Or was there a requirement that they have the same eigenvalues?

glossy crag
glossy crag
agile burrow
#

yeah same eigenvalues would imply similar because doing the change of basis would just give the same matrices, right?

glossy crag
#

In the words of our prophet Keith Conrad (PBUH) "We are interested in conditions that make a finite set of linear operators simultaneously diagonalizable: there is a basis in which the matrix representation of each operator is diagonal."

glossy crag
tender bough
glossy crag
#

Because EV for 1 operator => it's in one of the direct summands => all the operator are scalars on this summand => EV for all other operators as well.

tender bough
agile burrow
#

yes good work, I agree

#

sorry i was trolling earlier bc i am hungry

next obsidian
restive birch
#

really stuck on 10a here- any tips?

image.png
solar glacier
restive birch
#

awesome, thanks

glossy crag
#

Was just thinking about this, let {x,y,z} be a basis of V, let f be diagonal 1,2,2 and g diagonal 0,-1,3 (both wrpt this basis), then y+z is a 2-EV of f, but it's not anything for g.

glossy crag
#

Does anyone know what he's talking about here? Do you know whatsoever of any specific degree 2 subfields of C besides R, never mind being realclosed? Are these some AoC shenanigans?

image.png
hidden haven
#

The proof of the primitive element theorem gives the precise condition on a

rustic crown
#

the element cbrt(2) + a * omega should have degree 6, so you want it to have 6 different conjugates by separability

#

call the set of embeddings of K into some algebraically closed field S

#

then you have the polynomial f(x) = product_{s, t in S} {(s(cbrt(2)) - t(cbrt(2)) + x(s(omega) - t(omega))}

#

where the product is taken over s not equal to t

#

(or if you want, for every unordered pair {s, t} with distinct s, t)

#

f(a) = 0 if and only if cbrt(2)+a*omega doesn't have distinct conjugates

#

so only the roots of f are bad values of a

#

and now can you easily see what are the roots of f as its already completely factored

tender bough
rustic crown
#

that's an interesting question

rustic crown
#

its the other way

#

roots are x = -(s(cbrt2) - t(cbrt2))/(s(omega) - t(omega))

#

ofc if the denominator is 0, this won't be a valid root

#

conjugates of cbrt2 are cbrt2, cbrt2 * omega, cbrt2 * omega^2

#

and conjugates of omega are just omega and omega^2

#

mixing matching gives you all the s, t

#

it doesn't :p

#

i realized that later

#

at least for the case when you want to fuse together two generators into 1

#

but that strategy is nice

#

say you wanna fuse together 3 generators into one

#

you could actually use a + bx + cx^2

#

and again if you define f like above, you'll get that all but finitely many values of x will make this a generator

#

but if you iterate the condition for 2 generators, you only get the generator of the form a + bx + cy

#

idk where this is useful, but i find it nice eeveeKawaii

#

yea

#

true, there are like 6 choose 2 things

#

but you can convert it into a condition involving only the conjugates

#

like s(cbrt(2)) is just another conjugate of cbrt2

#

in general, if x is not -(alpha_i - alpha_j)/(beta_i' - beta_j') then you're happy

#

or something like this

#

one proof of the primitive element theorem avoids talking about the different ways to embed K, and directly asks you to pick x different from these finitely many values

#

it makes the proof shorter to write ig, i feel it makes it much harder to remember the idea

#

this is one of the first proofs of the theorem i read

image.png
#

but i don't like it much, to me it feels like a nice trick

rustic crown
#

if you follow what this proof is saying, then you just need to avoid 3 values of a

#

cbrt(2)/(omega - omega^2) * (1 - 1)
cbrt(2)/(omega - omega^2) * (omega - 1)
cbrt(2)/(omega - omega^2) * (omega^2 - 1)

#

the above thign should also just say the same after you only look at the elements which end up staying rational

#

like you could just apply s^-1 to the thing as it's already rational

#

and this will make it

  • (cbrt2 - T(cbrt2))/(omega - T(omega))
    where T is now s^-1t
#

now there are only 3 different values of T, as you want the denominator to stay non-zero

#

in the end it's the same computation

#

since we're done so much, lets finish it as well :p

#

so you don't want x to be
0
-cbrt(2) * omega^2
cbrt(2) * omega

#

oh and the last two are not rational

#

so you were right from the start

#

any non-zero rational works KEK

chilly ocean
#

Prove that the group $({2, 4, 6, 8}, \times \bmod{10})$ is cyclic

cloud walrusBOT
chilly ocean
#

for this, would it suffice to show that $2$ is a generator?

cloud walrusBOT
chilly ocean
#

because we kind of have $2^n \bmod{10}$

cloud walrusBOT
chilly ocean
#

i.e. $2 \times 2 \mod{10} = 4$, $2 \times 2 \times 2 \bmod{10} = 8$, $2^4 \bmod{10} = 6$

cloud walrusBOT
chilly ocean
#

gotcha

formal ermine
#

wait

#

where's your 1

chilly ocean
#

well 2^1 mod 10 = 2

formal ermine
#

your group has no neutral element

hot lake
#

It does have one

formal ermine
#

oh yeah mb lol

hot lake
#

But the 2^1 thing has nothing to do with it

chilly ocean
#

how would I go about finding four proper subgroups of a cyclic group of order 12?

#

I am not entirely sure about what constitutes a cyclic group of order 12

#

a group which has a generator with 12 elements?

formal ermine
cloud walrusBOT
#

yes yes yes no

391270706186420224.png
formal ermine
#

we say $g$ generates the group

cloud walrusBOT
#

yes yes yes no

391270706186420224.png
chilly ocean
#

so ${1, g, \hdots, g^{11}}$?

formal ermine
#

yes

#

g^12 = e

cloud walrusBOT
formal ermine
#

fwiw a cyclic group of order 12 only has 4 non trivial subgroups. iff G is cyclic then for every divisior d of |G|, there exists only one subgroup of order d

chilly ocean
#

right, because the order of the element is equal to the order of the cyclic subgroup

#

so since it's asking for "proper" subgroups I can't include the identity or the group itself

formal ermine
#

yes

chilly ocean
#

how do I go about finding them

formal ermine
#

you can look at for example $\bZ/12\bZ$, this group is $\Set{0,1,2,3,4,5,6,7,8,9,10,11}$ and a cyclic group of order 12

cloud walrusBOT
#

yes yes yes no

391270706186420224.png
chilly ocean
#

thanks, i'll try that

chilly ocean
#

I need to show that for each $n \in \mathbb{N}, \mathbb{Q} / \mathbb{Z}$ has a unique subgroup with order n. I found that $\frac{1}{n} \in \mathbb{Q} / \mathbb{Z}$ generates a subgroup $U$ of order n. I'm struggling a bit to show the uniqueness. I've tried assuming that there exists another subgroup V of order n, taking $x = \frac{p}{q} \in V$, then somehow showing with Lagrange's theorem that x must lie in U. I don't know whether this is the wrong approach or whether I am just too dumb to finish this off.

cloud walrusBOT
#

MatrixMaker

917072018602590248.png
chilly ocean
#

AH, maybe if I say that $ord(V) = n \implies \exists v \in V: ord(v) = n$

cloud walrusBOT
#

MatrixMaker

917072018602590248.png
chilly ocean
#

but that's not true in general, ah

ember field
chilly ocean
#

thanks, I will try showing that it is cyclic first

formal ermine
#

what does the set of torsion elements tell us about the module

rotund aurora
#

it is very explicit

#

well

#

In itself, the set of torsion elements is already interesting

#

But it is a submodule and a direct summand of the module M

#

In some cases

tender bough
#

If a set of linear operators is simultaneous diagonalizable, I wonder if \textit{the set of simultaneous eigenvectors} is unique.
Rigorously:
Given a set of simultaneously diagonalizable endomorphisms on a finite dimensional vector space V, the definition of simultaneous diagonalizability guarantees the existence of a set of eigenvalues corresponding to a basis consisting of simultaneous eigenvectors. Now for each eigenvalue $\lambda$ denote the eigenspace $E_\lambda$. Then is the disjoint union $\bigsqcup E_\lambda\setminus {0}$ all the simultaneous eigenvectors? i.e. Is it possible to have another vector not in the disjoint union, such that the vector is also an eigenvector of all the linear operators?

cloud walrusBOT
#

Mattuwu

336361466376617986.png
south patrol
#

Well, you're gonna need some conditions on the operators definitely

#

You can have trivial examples like taking 10 different multiples of the identity, since any basis is a simultaneous eigenbasis

#

Then you can make that slightly less trivial by using block matrices which are all multiples of the identity on some subspace, say, to one again create multiple different simultaneous eigenbases

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And so on

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Maybe I've misunderstood what you're saying though

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like I'm not sure what you mean by "the eigenspace", when all the operators will (presumably) have different eigenspaces in general

tender bough
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ohhh, true...

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I need to rethink about my question

slender hamlet
molten viper
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what's gonna be my best way of saying these 4 cosets are isomorphic toZ4 when taken as a group?

image.png
quiet pelican
slender hamlet
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Ok so for finding real closed non-isomorphic subfields of C, it should suffice to find two elements a, b of order 2 in Aut(C/Q) which are not conjugate

molten viper
chilly ocean
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Right now, I'm trying to assume that U is any subgroup of order n, then I pick $a = min{q \in (0, 1]: q + \mathbb{Z} \in U}$

cloud walrusBOT
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MatrixMaker

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chilly ocean
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then a should be the generating element of U, but I don't know how to proceed showing that.

quiet pelican
next obsidian
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Show that some power of an element of order n is of the form [1/n] in Q/Z

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Since this has order n this show all subgroups of order n are generated by [1/n] and are equal

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[] here refers to the equivalence class in the quotient

rotund aurora
chilly ocean
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Omg, I did it. Thank you @quiet pelican @next obsidian @ember field

glossy crag
glossy crag
glossy crag
# chilly ocean Omg, I did it. Thank you <@900064040536129616> <@146419489083228160> <@786879407...

Late to the party, but you could do it like this: first show that if x is an element of Q/Z and you write it as a fraction m/n in [0,1) with m&n coprime, then the order of x is n. This can be used to show that \langle 1/n \rangle is the unique cyclic subgroup of order n. If H is now your arbitrary subgroup of order n, bring all the elements of H to a common denominator a, say. This embeds H in the subgroup \langle 1/a \rangle, making H cyclic (as a subgroup of a cyclic group). But H has order n and is cyclic, so H=\langle 1/n \rangle.

chilly radish
rotund aurora
warm wyvern
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Also, that's interesting and unsatisfying at the same time angerysad

chilly radish
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Wdym

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Conjugation refers to i -> -i

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Trivial automorphism is just the identity

warm wyvern
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Oh, yeah ofc

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Why does that conjugation and aga^-1 type of conjugation have the same name?

south patrol
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Eh don't think there's any good reason lol

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Well there can't really be one due to C being abelian as a group

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But usually being conjugate implies some sort of symmetry between things which is definitely the case here

warm wyvern
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I see holothink

chilly ocean
slender hamlet
slender hamlet
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Ok wait this is the same as producing a real closed field K of cardinality same as R, such that K is not isomorphic to R

slender hamlet
slender hamlet
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Taking K to be the hyperreals works

main needle
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How to find the normalizer of subgroup of upper triangular matrices with 1's in the diagonals inside GL_3(Fp)

chilly ocean
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My homework says "Prove $A_n$ is normal"

cloud walrusBOT
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Iced Sugar

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chilly ocean
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what is An?

agile burrow
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A_n usually refers to the alternating group on a set of n elements, or the set of even permutations of a set of n elements. The question seems a bit incomplete, as one refers to a subgroup as being normal. I assume the full question is to show that A_n is a normal subgroup of S_n, the symmetric group on a set of n elements.

chilly ocean
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oh yes

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that is what the real question says

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yeah ima have to study that because idk what the set of permutations of a set even is

agile burrow
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ah yeah, it would be helpful to understand what the symmetric group is first. One way to think about it is as the collection of bijections from a set to itself, with the group operation being composition

obtuse bear
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there's a general theorem that if H is a subgroup of G with index 2, then H is normal in G. You could apply that here if you know the orders of A_n and S_n

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another method is to recognize that if you conjugate an element of S_n by another element of S_n, the cycle structure is preserved

woeful sage
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is an exterior algebra this? $$E = \bigoplus_{k = 0}^n \bigwedge\nolimits^k V$$

cloud walrusBOT
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Neamesis

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woeful sage
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in the wiki they've said stuff about quotient of a tensor algebra or smth i dont get that bhappy

elder wave
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it is

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you obtain this from the tensor algebra by taking a quotient wrt. to v+w \otimes w+v because of alternating stuff

south patrol
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I think the point is basically just that like

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You can either quotient stuff out from tensor products and assemble into the exterior algebra, or you can quotient stuff out from the tensor algebra first and then recover exterior powers from the grading

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Fortunately the construction isn't very involved so it is fairly clear these r equivalent

median pawn
woeful sage
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oh hey a tensor algebra is like a fock space (or rather i guess the fock space is a tensor algebra) catThin4K

median pawn
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oops sorry for interrupting