groups-rings-fields

[coral shale]

This is also a combinatorics exercise - you want to find the different ways of chopping 4 items into groups

not for just the shape

You only care about this when it comes to order

(I say 'shape' theres probably an actual term ive forgotten)

Conjugacy class or something

[chilly ocean]

and this'll help me understand there's nothing of order 12?

I think it kinda makes sense

[coral shale]

yes, find the orders of each of these

[chilly ocean]

lcm((a)(b)(c)(d)) = lcm(1, 1, 1, 1) = 1

lcm((a b c)) = 3

lcm((a b c d)) = 4

lcm((a b)(c d)) = 2

hm

[coral shale]

btw we usually omit cycles of order 1 in decompositions, but either choose to include them or dont

If you include them, (abc) = (abc)(d)

[chilly ocean]

right

[coral shale]

(ab) = (ab)(c)(d)

[chilly ocean]

understood

and this makes no difference when calculating lcm

because it's just one

[coral shale]

yes, am just nitpicking this line

[chilly ocean]

but have we listed all the possible "shapes" of the permutations

[coral shale]

.

Sorry to be more precise

its the problem of splitting 4 into a sum of 4 positive numbers

4 = 1+1+1+1
4 = 2+1+1
...

[coral shale]

this is equivalent. You can solve this systematically to convince yourself you have

[chilly ocean]

I see

thank you @coral shale

I appreciate all the help

I think I've got it

oh

@coral shale I understand why you wanted me to observe the "shapes" of these permutations

(sorry for the ping)

[coral shale]

[chilly ocean]

wait

ill send a more precise screenshot

[coral shale]

Conjugacy class is the proper term

[coral shale]

(it turns out all permutations of the same 'shape' are equal up to conjugation)

[coral shale]

yh. You can try the same exercise for say S_6

its fairly straightforward to do

[chilly ocean]

but why is the order of (a_1 a_2 a_3 . . . a_n) (assuming they a_j != a_k) always going to be n

[coral shale]

So think of a permutation as a function

f = (123...n)

If you apply f repeatedly to 1

ffffff(1)

If you do this less than n times, it can never map to 1

[chilly ocean]

ohhhhhhhhhhhhhhhhhhhhhhhhh

[coral shale]

f^k(1) = 1 + k

same argument for the other elements

[chilly ocean]

are you a graduate student?

[coral shale]

i am

[chilly ocean]

cool

[coral shale]

or at least

[chilly ocean]

anyway thanks again

[coral shale]

i graduated

[chilly ocean]

oh

[coral shale]

im not a student any more

[chilly ocean]

haha

well congrats

[coral shale]

i might be a student again in the future 👋

[rustic crown]

ooh what you doing now shuwi

[coral shale]

work

quant

[rustic crown]

oh

[coral shale]

i cant wait to do more real math

[rustic crown]

i thought some top secret research >.<

[coral shale]

but no, finance math is somewhat interesting

i have been intrigued so far

[rustic crown]

tell more

[coral shale]

all a bunch of calculus and probability lol

[rustic crown]

no alg?

[coral shale]

nup

[rustic crown]

[coral shale]

there will be linear algebra when i learn ML stuff ig

[rustic crown]

[coral shale]

dw when i get more into it im sure i can shove some algebra in

[rustic crown]

[warm wyvern]

Nope

[coral shale]

that was a statement

dont lie to me dq

[warm wyvern]

I've never seen something more advanced than eigen vectors from ML

(wtf did i type )

[coral shale]

if u so say

good news for me

because i basically didnt get any further than eigen ever

[warm wyvern]

You might wanna learn about optimization tho

Which sounds dope af

[feral agate]

How would one go about proving that endomorphism of (V) is isomorphic to the 1-1 tensor, namely, T: V x V* -> K, where V is a vector space and K the underlying field

[rotund aurora]

So you want to show V x V^* = L(V) right

[south patrol]

I assume you mean the spaces of such maps?

Probably worth showing more generally that there's an isomorphism Hom(V,W) with V* x W

[rotund aurora]

yeah

there is one map that should immediately come to mind

you require finite dimensionality of all vector spaces I think tho

otherwise you just get an injection (which is particularly nice)

[tribal moss]

Yes, you only get an isomorphism in finite dimension (otherwise there's no tensor representing the identity map on V).

[novel plover]

"Give an example of an abelian group G that has exactly 1000 elements" ? ?

[rotund aurora]

Z/1000Z

[glossy crag]

I'll do you one better: (Z/2Z)^{500}

[novel plover]

Wait what's Z/2Z again? I forgot

[glossy crag]

Same thing Z/1000Z is

[coral shale]

integers mod 2

[coral shale]

heck what?

2^500

[tribal moss]

That has a hell of a lot more than 1000 elements.

[sacred pier]

What is the meaning of equation 2? Thank you

[coral shale]

inner product?

[glossy crag]

"I'm good at mental arithmetic" 🤦‍♂️

[novel plover]

So the 1000 elements are the 1000 congruence classes. for example all the integers divisible by 1, by 2, .... by 1000

[coral shale]

no.

you're dividing by the same thing .

1000

You're interested in the remainder

[rustic crown]

am i?

[coral shale]

ur not

[glossy crag]

First one yes, second one no. All integers leaving remainder 0 when divided by 1000, leaving remainder 1 when divided by 1000 and so on till 999

[rustic crown]

[novel plover]

Ohh okay. Nice. Are there any other groups with exactly 1000 elements?

[coral spindle]

Yeah

[coral shale]

most likely

(C_10)^3

[rustic crown]

@coral spindle that story was really fun. ping me if you ever do that again

[rotund aurora]

[coral spindle]

There's a theorem that classifies all Ableian groups of a particular order

[tribal moss]

Is it more readable as $\langle a, P b\rangle = \langle a_W, P b\rangle = \langle P a, b\rangle$?

[coral spindle]

Lmao it is something I do from time to time, I'll @ you if I do

[novel plover]

Fundamental theorem of finite abelian groups?

[coral spindle]

finitely generated, but yes that's right

[coral shale]

latex's u's and v's smh

[coral spindle]

In fact I'll give you an example of a group of order 1000 that isn't Z/1000Z

Z/2Z x Z/2Z x Z/250Z

[novel plover]

Oh it has to be abelian too, I forgot that part

[rotund aurora]

the description of finite abelian groups is purely arithmetical (see the structure theorem). If you drop abelianity then the situation is much more wild. https://groupprops.subwiki.org/wiki/Groups_of_order_1000#:~:text=gap> SmallGroupsInformation(1000)%3B There,199 groups of order 1000.

[coral spindle]

Yes, asking for all the groups of order 1000 is a much harder question!

[sacred pier]

I am not familiar with either. If that is inner product I have a direction on it. Thank you

[tribal moss]

Fair point, I switched to a and b.

[cloud walrus]

Troposphere

[rotund aurora]

there are 199 groups of order 1000, as says the link provided

[coral shale]

|A x B| = |A||B|

use this fact

[novel plover]

199 non-abelian groups?

[coral shale]

and u can easily make groups of order 1000

[tribal moss]

Unless I'm mistaken, there should be only 9 abelian ones.

[rustic crown]

p(3)^2?

right

[rotund aurora]

the other day I thought that it might be possible to give asymptotics on the number of isomorphism classes of groups of order n

[tribal moss]

If p counts unordered partitions, then yes.

[novel plover]

The group U_1000 of solutions of z^1000 = 1 in C under multiplication is the example my book gave of an abelian group with 1000 elements

[rustic crown]

(ig 199 included both, abelian and non-abelian)

[glossy crag]

It's the same thing as Z/1000Z (group isomorphism).

[coral shale]

Remind me whats the group U_n smh

I never recall ever seeing this in courses

[glossy crag]

Presumably the 1000th roots of unity.

[rustic crown]

i thought U_n was used for (Z/nZ)*

[novel plover]

It's the set {z : z^n = 1} where z is a complex number

[coral shale]

is this common notation

[rustic crown]

common enough that people don't complain when they want to be lazy

[glossy crag]

Ireland/Rosen use it in their book.

[coral shale]

i complain 😶

but alright

[coral shale]

surely this

[coral shale]

and this

dont coincide do they

|{z : z^n = 1}| = n
|(Z/nZ)*| = phi(n)

[glossy crag]

The latter has 400 elements.

[rustic crown]

yee

[coral shale]

well this is a moment

[novel plover]

U_n under multiiplication is isomorphic to Z_n with addition mod n

[coral shale]

|{z : z^n = 1}| = n is just C_n or Z_n. We don't need yet another letter for this thanks

[coral shale]

I'll take this

[rustic crown]

[tribal moss]

Z/nZ as a additive group.

[coral shale]

huh wheres the mistake

[tribal moss]

Whoops, I see what you were reacting to now.

Never mind me.

[rustic crown]

i think mu_n is a notation i've seen for n-th roots of unity

but that was in the context of group schemes

so ig i should C-valued points of that

[delicate bloom]

I've seen this notation before too, and since I don't know what group schemes are, it must have been in another context

[novel plover]

Does the order in which you state the axioms of a group matter?

[delicate bloom]

i hope not

[coral shale]

You should state identity before inverse

[novel plover]

Ah yeah, good point

[toxic zephyr]

i can only imagine an extreme pedant making a fuss about it.

[coral shale]

Im not being a pedant

[toxic zephyr]

no i didn't mean you or what you said. what you said is just good advice.

[delicate bloom]

since there are programming languages that allow you to do things with variables that aren't defined until later down, I feel like it doesn't matter imo

[coral shale]

js moment

[delicate bloom]

but I like the convenience of identity before inverse, sorta stupid to not do it that way, you're right

[rustic crown]

depends on how you define a group. if you define it as a quadruple (G, m, e, i) where m : G x G --> G, e : 1 --> G and i : G --> G, then we already know what e is :3

[coral shale]

think this is a bit better

in terms of some well-definedness practicalities, perhaps

[novel plover]

How many possible binary operations are there on a set of 2 elements, and how many of these give a group structure?

[coral spindle]

Do you want hints with this or do you just want an answer

[novel plover]

Yeah maybe some hints to help me understand it better. I feel like there should be 8 possible binary operations. Not sure how many would be groups, but maybe you can tell from making a table

[coral shale]

take the set defn of function

[coral spindle]

Hint: the number of functions from a set A to a set B is |B|^|A|.

Hint: there is only one group of order 2, up to isomorphism.

[wooden ember]

I just looked it up

There are 199

[novel plover]

So A = {a,b} and B = {a,b,e, a^-1} ? 4^2?

[coral spindle]

I don't like guesswork. I'd rather see a justification.

[coral shale]

define what a binary operation on G is

[novel plover]

Because a binary structure is taking a pair (a,b) and assigning it to an element

[coral shale]

pair from where

element from where

[novel plover]

You have your set S with two elements {a,b} , the binary structure is mapping SxS -> S ah so |A| = 4 and |B| = 2

[coral spindle]

So how many such operations are there?

[coral shale]

So to be explicit

A = SxS

B = S

Or really, just write f : G x G -> G

[novel plover]

So, SxS can mean (a,a) (a,b) (b,a) or (b,b)

[coral spindle]

No

[coral shale]

not mean

has the elements

S x S = {what you listed}

[coral spindle]

S x S means a particular set – it contains those elements

It's really important that you don't confuse these concepts!

[novel plover]

Ah, so SxS = {(a,a) (a,b) (b,a) (b,b)}

[coral shale]

yes and S = {a, b}

[novel plover]

And each of those gets map to either a or b

[coral shale]

anyways, so using this . .

[novel plover]

Yep, there are 4 things with 2 possibilities for each so 2^4

[coral spindle]

Good job

[novel plover]

And then you hinted maybe only one of these gives a group structure. 1/16?

[coral shale]

at least 1.

Dont guess.

[coral spindle]

Again, I don't like guesswork.

If you don't have a watertight justification, you haven't answered the question

[coral shale]

The way I'd approach this problem is Cayley table

but not necessary

[novel plover]

I guess associativity places a heavy restraint on what the Cayley table can look like

For 2 elements

[coral shale]

no

associativity is usually hard to detect on a cayley table

There are 16 binary operations, but quite a few you can immediately reject from group axioms or immediate properties that follow

[coral spindle]

I would strongly recommend using the fact that we know what the group of order two looks like already.

But if you want to do it from scratch, then yeah, follow mu's advice

[novel plover]

Group of order 2 looks like {e,a} so a&e=e&a = a, and e&e=e, and it must have inverse so a&a=e

[coral shale]

drawing the cayley table for this would help, too

Yeah, then use this to determine which binary operations are possible

[delicate bloom]

one way I have in mind is use the fact that multiplication by an element of the group is a bijection

[coral spindle]

Big hint, imo: ||If you know that the set is a group of order 2, then if you know which element is the identity, then you know the entire structure of the group.||

[novel plover]

I see. I think I understand. Thanks

[coral spindle]

It was spoilered for a reason >:(

[novel plover]

So for a set with 3 elements I'm thinking there's 19683 possible binary operations

[coral spindle]

Are you trying to do this again with order 3 groups? I don't think you've finished the order 2 case

Also, this is going to get a lot harder

[coral shale]

nowhere near.

[coral spindle]

Nah that's correct; 3^9 = 19683

[tender wharf]

why is there guessing

[coral shale]

oh nvm

[tender wharf]

order 3 group is cyclic

[coral shale]

I think the fastest way to compute for arbitrary order is to list possible group structures and take permutations.

taking permutations is not necessarily simple, has to be thought (some may result in the same binary operation)

[coral spindle]

Well, if you have representatives G_1, ..., G_m of the groups of order n up to isomorphism

You can count the number of times G_i occurs as a binary operation: it's |Sym(n)/Aut(G_i)| = n! / |Aut(G_i)|

This actually gives away the answer for the case that n is a prime 🙊

Naughty me

[tribal moss]

I think even the 4-element and 5-element cases can probably just be winged, knowing all the possible group structures.

[coral spindle]

I daresay if you want to calculate this in a human lifetime for, idk, n > 10, you're going to have to know all the groups of that order

[tribal moss]

And the symmetries of each of them.

[coral spindle]

Fwiw we have enough theory to make that a fairly easy task for certain orders

That too

[coral shale]

thats very nice and makes sense

You write Sym(n)/Aut(G_i)

hmmm

[coral spindle]

You can see this via the action of Sym(n) on the set of binary operations on a set with n elements

[coral shale]

ill have a think. ty

[coral spindle]

It should come from orb-stab, I reckon

[tribal moss]

That's nice as a general theory, but I'm not sure we should discard ad-hoc reasoning quite that soon. For example, how many binary operations on {a,b,c,d} give us V_4? It seems faster to say "well, all we can really choose is which element is the identity" than it is to compute the size of Aut(V_4).

[coral spindle]

This is kinda the theoryified version of that question

[coral shale]

so |Aut(V_4)| is 6, if I follow

[coral spindle]

Aut(G) in my head allows us to sort out the information that we need to specify the group

[coral shale]

3!, of course.

[coral spindle]

so for the spoilered hint I gave, this follows from Aut(Z/2Z) = 1

We do need to choose a bit more than the identity in the case of V_4, as mu's calculation points out

We need to choose two generators, and we're done

but yeah

Sym chooses the identity, and Aut chooses the generators

[tribal moss]

No, the group operation doesn't actually depend on which elements we call generators.

The product of any two non-identity elements is the third non-identity element.

[coral spindle]

Oh mb I had it the wrong way in my head

Yeah you're right, it's quite the opposite

because Aut(V_4) is 3!, it's entirely dependent on the identity

my mistake

[coral shale]

A lot of these counting problems in aa seem to boil down to combinatorics thinks

[coral spindle]

For some reason I was fixating on it being equal to 1, not (n-1)!

[long geyser]

counting does indeed seem to boil down to counting

[coral shale]

indeed.

[tribal moss]

Who'd have thought counting problems would boil down to combinatorics? :trollface:

[coral shale]

thats more combinatorics than most of us would like tho

[novel plover]

Yea lol

[coral spindle]

Anyway, this general answer gives us the prime case very nicely

[coral shale]

iirc |Aut(G)| is non-trivial right?

i literally have no picture in head rn hm

[tribal moss]

For an arbitrary group it can be difficult, yes.

[coral shale]

ok in prime case, just generator map

|Aut(C_p)| = ... p

i think

[coral spindle]

Not quite

[coral shale]

p-1

[coral spindle]

Yeah that's right

[coral shale]

I never really got comfortable with automorphism stuff

[tribal moss]

E.g. the exceptional outer automorphism of S_6 is not something one would just guess from staring at a systematic description of the group.

[coral spindle]

Tbh, it's really hard to guess almost any property of a group from a description of it

Something something Adian–Rabin

[tribal moss]

Finite groups are hard, let's go shopping!

[coral shale]

I'm slowly clocking that same binary operation iff automorphism rn

[delicate bloom]

just want to remark the notation is particularly nice $|S^{S\times S}|=|S|^{|S|\times|S|}$

[cloud walrus]

Merosity

[tribal moss]

So for n=10 there's exactly a googol different binary operations.

[delicate bloom]

damn, an actual application of googol, first for everything

time to frantically work out what gives a googolplex now

[coral spindle]

And there are only 2 groups of order 10

[coral shale]

D5, C2C5, C10 at least

[coral spindle]

We can probably work out what proportion of these are groups now, let me just think

[rustic crown]

2 and 3 are same

[coral spindle]

C2 x C5 = C10 :)

[coral shale]

shh

[coral spindle]

Righty, so

|Aut(C_10)| = phi(10) = 4,
|Aut(D_10)| = 5*phi(5) = 20, after I looked it up lol

So in total we have 10!/4 + 10!/20 groups

I like D_2n notation

[delicate bloom]

,w (10!/4 + 10!/20)/googol

[coral spindle]

that's 1088640 groups out of googol

[rustic crown]

lol i asked it in a weird way

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Merosity#0135

[coral spindle]

so about 10^6/10^100

Extremely small number

yup as Wolfram says, about 10^-94

[rustic crown]

do you like S_n! then

[coral spindle]

Nah

But I write Sym(n) and Dih(2n) anyway

I think of Sym as a group on the set n = {0, ..., n-1}

(acting on the set, I mean)

[delicate bloom]

i guess the funny factoid I'm capable of remembering to spit off to kids is, "despite there being a googol of binary operations on 10 elements, there are only two groups of order 10"

[rustic crown]

write that 2 in binary

[coral spindle]

If you ever write a book on intro group theory lol

[delicate bloom]

but it's shoving the symmetries under the rug there

[tribal moss]

Apples to, um, some kind of fruit.

[coral spindle]

Well I mean, the 10^-94 is still a cool number to throw around

And you could give this as an exercise!

[delicate bloom]

I just want an excuse to say googol

[tribal moss]

Yeah "only a million or so" kind of spoils the rhetorical oomph.

[coral spindle]

But a googol is so big dammit!

Okokok

you could say

[tribal moss]

About 17 times larger than a million.

[delicate bloom]

hah

[rustic crown]

numbers bigger than 20 are too big

[coral spindle]

"For every group structure on 10 elements, there are about 10^94 binary operations on that set that don't form a group"

This one is pretty impressive

That's like, almost a googol

[delicate bloom]

"For every group structure on 11 elements, there are over a googol binary operations on that set that don't form a group"

[tribal moss]

For every field structure on 10 elements there are infinitely many pairs of binary operations that don't form a field.

Somehow doesn't sound quite as impressive.

[delicate bloom]

lol

[rustic crown]

say it for 11 then

[coral spindle]

For 11 it's boring smh

and you can't say googol!

[delicate bloom]

10 is a power of a prime, p^{log_p(10)} for all primes p!

[rotund aurora]

googolplex

[coral shale]

Oh I clocked it. By definition automorphism not only preserves set, but the group operation as well. So by definition the binary operation is invariant under automorphism

This means isomorphic groups are automorphic iff they have the same underlying set and binary operation

[tribal moss]

Congratulations on the insight. (It's trivial once you see it).

[coral shale]

I kept trying to think about the homomorphism defn and getting nowhere

[coral spindle]

Once you've got the right definitions down it pops out I think

[white oxide]

how does it follow that there is only one subgroup of Zn of order n/d?

[tribal moss]

Because that single subgroup contains all the elements whose order divides n/d.

[barren sierra]

How would I find an example sequence that under F isn't left-exact for the last part of the question

[south patrol]

You can take A,B to be Z, for example

[barren sierra]

I can?

Hm I tried that but I thought that didn't work

lemme check my work

[south patrol]

What map did you use

There is a clear / fairyl natural one imo

[barren sierra]

identity

[south patrol]

No

The identity will always be sent to identity for any functor

[barren sierra]

yea

[south patrol]

Think of another choice of map

[barren sierra]

yea idk what else I could use

Oh

map z to 2z

I think

yea that works

[south patrol]

Indeed

owo

[barren sierra]

and then I think C should be Z / 2Z

or does C = 0 work, that may be easier

no C = 0 doesn't work

[oblique river]

it doesn't matter what C is

[barren sierra]

but Z / 2Z maps to "itself" under the functor so it's cool

[oblique river]

C is just the quotient of B by im(A)

[barren sierra]

??

oh

[oblique river]

the only thing that matters is that you have an injective A --> B which is no longer injective

after appyling the functor

[barren sierra]

yea

[tender wharf]

mocha

[toxic zephyr]

so the first one makes sense to me. b1b2^-1 is still going to be a string of elements from A with exponents either 1 or -1. but i'm not sure how to go about adjusting this to make it normal... the thing i'm really unsure about is how to use the elements of A to make sure that conjugation by an arbitrary element of G stays in the set.

would it just be something like all elements of the form $\prod_{i=1}^n g_i\inv a_ig_i$ for some $n\in\bN$?

[cloud walrus]

nilpotent nix

[toxic zephyr]

btw just looking for a hint, not a solution 😅

[toxic zephyr]

yeah this doesn't work

[ionic vale]

H = <A> is gonna be a subgroup of the desired normal subgroup (calling it N). so in particular gHg^-1 is also gonna need to be a subgroup of N, for each g. so what does that tell you about what elements have to be in N?

they'll be ||g * a_1^e_1 * a_2^e_2 * ... * a_n^e_n * g^-1, for each g in G||

and then these will be enough to form a normal subgroup

[toxic zephyr]

so i thought about that. but i don't see how it will be closed under multiplication, since it will have a g1^-1g2 in the middle and they aren't necessarily the same g...

[ionic vale]

oh true 😅

mb

[toxic zephyr]

sorry for the low quality screenshot but these are the exercises my prof listed as things to look at related to this one

it's weird to me that it seems easier when there are two operations lol. i guess maybe because one is always commutative

[ionic vale]

i shouldn't have begun to think about this at 3am. but well

[ionic vale]

it'll have to at least contain the subgroup generated by these guys, which looks pretty ugly, so it seems like something that'll look pretty ugly....

ohhh, okay, so

the subgroup generated by the gHg^-1 actually works

though you need a little trick to show it's normal

say $N = \langle gHg^{-1},\text{ for each $g\in G$}\rangle$. then each $n\in N$ will be a product like $n = \prod g_ih_ig_i^{-1}$. when we try to take the conjugate of this, what seems to impede us from getting an element of $N$ again are the junctions like $g_i^{-1}g_{i+1}$. is there something smart we can do at those junctions to circumvent the issue?

[cloud walrus]

lambdcalculus

[ionic vale]

lil' extra: ||the trick will let us turn gng^-1 into a product of elements of N||

the actual trick: ||g_i^-1 * g_{i+1} = g_i^-1 * g^-1 * g * g_{i+1} = (gg_i)^-1 * gg_{i+1}||

having to type without latex so the bot doesn't show stuff is pain

i assume there's a way to make it hide the output as a spoiler. anyone know how one does that?

[toxic zephyr]

ahahhh i see. so the original thing i thought does actually work! since we can insert gg^-1's between every term, conjugating the product by g just turns it into a product of (g_ig)^-1 ai (g_ig)

[tender wharf]

,help

[cloud walrus]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[toxic zephyr]

@ionic vale thank you so much for all your help even though it's past 3am for you 🙏

[tender wharf]

I think ,texconfig should have an option to spoiler the output

,texconfig

[cloud walrus]

Personal LaTeX configuration.

​ ​ ​ ​ ​ ​ autotex: Disabled (may be overriden by guild settings)
keepsourcefor: Don't delete source (may be overriden by the guild)
​ ​ ​ ​ ​ ​ ​ colour: Using the white colourscheme
​ ​ ​ alwaysmath: Disabled
​ ​ ​ alwayswide: Disabled
​ ​ ​ ​ namestyle: HIDDEN
autotex_level: WEAK

To see more detailed information use ,texconfig <option>.
To set an option use ,texconfig <option> <value>.

Preamble

Using a custom personal preamble with 203 lines!
Use ,preamble to view or modify your preamble!

Feb 5, 2023, 7:13 AM

[ionic vale]

i figured it out

,texsp $boop$

[tender wharf]

you mean texsp

[ionic vale]

yrp

[tender wharf]

ah yeah

but you need to turn keepsource to like 0

[ionic vale]

i mean, if you spoil the source as well it's fine. though if i recall it'll do some weird thing

,texsp ||$boop$||

[cloud walrus]

lambdcalculus

[tender wharf]

,texsp ||u||

[tender wharf]

yeah that darn thing

[ionic vale]

no worries.

[somber sleet]

hey guys, I'm back. How do you intuitively see if this field extensions is simple or not? When you get it, how do you prove your claim?

[rustic crown]

finite+separable => simple

That's not from intuition tho lol

[somber sleet]

what is your definition of separable? I don't have it in my skript

also I wanted to ask, for a,b in L a field extension of K, does K(a,b)=K(a)K(b) always hold?

[rustic crown]

there are many equivalent definitions, you say a polynomial is separable if it doesnt' have repeated roots. and then you say an algebraic extension F/k is separable if minimal polynomial of each a in F over k is separable.

[somber sleet]

oh yeah, now I know what you mean, sorry my german and english are kinda confused sometimes haha

[rustic crown]

what do you mean by * here?

[somber sleet]

I meant the "multiplication"

but I'm confused too

cause I saw it in some solution, but I think they mean the generated "Zwischenkörper"(I have to look the word in english up) by K(a) and K(b)

[rustic crown]

if E and F are subfields of some larger field L, then EF is defined as the smallest subfield containing E and F

which is also called the subfield generated by E and F

[somber sleet]

is defined as the union right?

[rustic crown]

and EF = {e1f1 + ... + enfn | ei in E, fi in F}

[somber sleet]

I don't know, maybe you see something here

I think we are both on the same page

[rustic crown]

idk german

[rustic crown]

(but this is only true if like L/k is like algebraic or something, otherwise you would also need denominators)

so yea E*F isn't exactly just {ef | e in E and f in F}

[somber sleet]

maybe you recognize some symbols

[rustic crown]

looks like they're saying K1K2 is the subfield gneerated by the the union?

[somber sleet]

yeah, I mean, for this to hold E/L hast to be finite for it to hold

[somber sleet]

yes exactly

[rustic crown]

any field extension of char 0 fields would automatically be separable, so here all finite extensions are simple

[somber sleet]

so we get "problems" when charK=p for a prime number right?

[somber sleet]

this means that the smallest subfield of K is Q right?

[rustic crown]

yep, the standard example for a non-simple extension is this
F = F_p(s, t) and k = F_p(s^p, t^p)
so [F:k] = p^2 but this has inifnitely many intermediate fields (which doesn't happen for simple extensions), the ones generated by s + a*t and a in k.

[rustic crown]

yep, char 0 fields contain a copy of Q

[formal ermine]

intermediate field

[rustic crown]

ah is this like that chess move

zwischenzug somthing

which means an in-between move

[formal ermine]

zwischen = between

[rustic crown]

det knows another german word now

[formal ermine]

primitive element theorem my beloved

[formal ermine]

did you mean K(a)(b) here?

[rustic crown]

if E/k and F/k are algebraic extensions, then so is EF/k and in this case EF is finite sums of elementary products e*f

(ofc to define EF, you assume the are subfields of some large enough field)

that's why we use the notation EF

but the definition is the subfield generated by E and F

K(a) * K(b) = K(a, b) = K(a)(b)

[rotund aurora]

Just to make sure I'm not doing anything stupid, (a,b)^2=(a^2,ab,b^2), as ideals

[rustic crown]

yep

(assuming everything is commutative)

[formal ermine]

are there very non commutative rings?

like where + isn't commutative either?

[rotund aurora]

Take a "complete" set of bijections with the operation of composition. Make the sum and multiplication operations the same, the composition

that works, but probably not what you were expecting xdd

[coral shale]

this is usually not considered

[coral shale]

talk about groups

[rustic crown]

it's not distributive

[coral shale]

a group is furthest from being commutative when its abelianization is the trivial set

[hot lake]

if you want distributivity it implies that + is commutative

[rotund aurora]

ahh you are right

[rustic crown]

wait really?

[coral shale]

cant u have left and right

[rustic crown]

oh now that you say it, i think i did this exercise sometime

[rotund aurora]

actually wasn't this asked once some time ago?

[hot lake]

like uh, you look at (1+a)(1+b), you can distribute it in 2 ways and then you compare what they give you

[coral shale]

ic

[rustic crown]

if we don't have a 1, then we could only say that the elements {a*b : a, b in R} commute with +, right?

[formal ermine]

I was more asking about a theory behind them lol

something like very non commutative algebra

[atomic mesa]

The book says I can’t solve this using group theory. I tried to see why. Does the problem happen because the group isn’t of prime order?

[formal ermine]

do you know what a splitting field is

[chilly radish]

I mean, you can absolutely prove this with group theory, but you need some more advanced tools

[atomic mesa]

Nope

[formal ermine]

[chilly radish]

you don't really need any field theory to prove this

[formal ermine]

hmm

[chilly radish]

you can do it using some basic sylow theory

[atomic mesa]

Sylow won’t be covered until next year…

[chilly radish]

I don't know any simpler proof. Elementary methods won't really help you here, at least not just lagrange on its own

[atomic mesa]

I ask this because I remembered something about groups of prime order being cyclic

I just wanted to check that my above working was correct

[formal ermine]

this won't help you here

[atomic mesa]

Apparently there’s a difference between F_p and Z/pZ

[formal ermine]

they're isomorphic

[chilly radish]

one of those usually dentoes a field

but it's only a semantic difference

[atomic mesa]

But by definitions I think F_p is like a world of integers that stops after p-1

Z/pZ has elements which are equivalence classes, so every integer falls into one class

[chilly radish]

Here's how to prove it with sylow: ||In general, you can prove any finite subgroup of the mult. group of a field is cyclic. What you do is, for every sylow subgroup, say of order p^k, note that x^p^k-1=0 has at most p^k roots, so there are at most p^k elements of order dividing p^k, but this implies the sylow subgroup is unique, hence it's normal. By similar reasoning, you can conclude that the sylow subgroup is cyclic. Why? Because there are at most p^(k-1) elements of order p^(k-1), but there are p^k elements in the sylow subgroup, so there must be some element in it that has order greater than p^(k-1) and dividing p^k, i.e. it must be of order p^k. Finally, normality implies that your group is a direct product of its sylow subgroups, and each of these is cyclic with coprime order, and you finish by the chinese remainder theorem||

[novel parrot]

does someone know why both a(x) and b(x) are forced to have constant term of 0 ?

i thought only of the constant terms is forced be 0 only

[chilly radish]

@formal ermine

[atomic mesa]

Yeah I was told this as well, but I thought they were literally the exact same thing, just a notational difference

[formal ermine]

the proof I had in mind is ||it's the splitting field of x^p^k - x so its mult group is U_(p^k - 1) which is cyclic||

[atomic mesa]

In response to this

[formal ermine]

fields are unique up to isomorphism

as in like there's only 1 field of order n up to iso.

if that's what you mean

[atomic mesa]

Oh I see

Btw if H is isomorphic to G, is G isomorphic to H?

[formal ermine]

yes

isomorphisms are bijections

[atomic mesa]

I think yes because elements are just relabelled but it doesn’t feel rigourous

[chilly radish]

finite* fields

[formal ermine]

"but what about the empty set"

[chilly radish]

This doesn't show F_p^\times is cyclic tho

I mean it's a very important distinction, there are infinitely many nonisomorphic infinite fields, so it's not true that fields are that nice

[formal ermine]

you can show that U_n is cyclic using the structure theorem

[chilly radish]

then you don't need fields a priori anyways...Like I said

the initial question was about F_p^\times

[formal ermine]

yes, I linked it to U_n

[chilly radish]

you can see by inspection that it's the same thing

[median pawn]

suppose L is a field extension of K

is it true that every element a of L is the root of some irreducible polynomial over K?

[formal ermine]

those are called algebraic extensions

or well

algebraic extensions are when every element of the extension field is a root of some polynomial over K

without the irreducible modifier here

and there are non algebraic extensions

[chilly radish]

I mean, it's equivalent

but yea these are exactly the algebraic extensions

[formal ermine]

right

[glossy crag]

There are elementary proofs as long as you know F_p is a field and that the polynomial x^{p-1}-1 has p-1 different roots in it (all lying in F*_p). All you need is this and some basic group theory (knowing how orders of elements work). Page 10 of https://kconrad.math.uconn.edu/blurbs/ugradnumthy/ordersmodm.pdf, so long as you know what polynomials over a field are, you should be able to understand this (this is the simplest and clearest proof I know). https://kconrad.math.uconn.edu/blurbs/grouptheory/cyclicmodp.pdf has a bunch of other proofs of this one fact, if you are curious to see more.

[median pawn]

interesting i see

is every finite extension algebraic?

[formal ermine]

yes

[median pawn]

why so?

that's really what my question is

[formal ermine]

because L/K is finite there exists an n such that { 1, x, ..., x^n } is linearly dependent over K

fwiw the converse only holds true when L/K is finitely generated

[south patrol]

Oh never seen a proof using Sylow theory

My method to show any finite subgroup of a field is cyclic would be ||use the fact that a group G of order n is cyclic iff it has <= phi(k) elements of order (exactly) k for each k|n and note there are <= k elements of order dividing k for each k|n giving the result after splitting up||

[chilly radish]

yea, that's pretty much the same thing innit

like i'd prove the thing you said with sylow theory also

you can do it elementarily too tbf

but it's more annoying

[south patrol]

I wouldn't use sylow theory for that

But ye

[chilly radish]

how would you prove it without it being a PITA

[formal ermine]

pita?

[south patrol]

|| sum of phi(d) over d|n is n just from considering Z/nZ or by other methods and then from that this is easy||

Ofc if I were asked in an exam or at gunpoint I'd just say structure theorem as above but ye that is overkill

[chilly radish]

pain in the arse

[formal ermine]

lol I thought it would be something like PID

[chilly radish]

yea that works but I don't like working with the totient function

lmao

[coral shale]

Where |G| = n, you write S_n/Aut(G). Is this a quotient group!? I don't see how Aut(G) is normal

Conjugation of an automorphism by any permutation which moves the identity will surely not yield an automorphism

[coral spindle]

Doesn't need to be

A/B is just the set of cosets, which certainly exists

[coral shale]

oh.

[coral spindle]

a more normal notation would be [Sym(n) : Aut(G)]

[south patrol]

Ye fair

[coral shale]

Yes this makes sense then yh

[somber sleet]

What is the best way to prove, that if L/K is an algebraic expansion, then a and b in L are algebraic over K iff a+b and ab is alg over K

[rustic crown]

you just need this one lemma: F/k is f.g. extension say F = k(a1, a2, ..., an) then TFAE

1. F/k is algebraic
2. each a_i is algebraic/k
3. F/k is finite

you probably have seen this by now

so if a, b are alg/K then K(a, b)/K is an alg extension, so a+b and ab are alg/K

conversely, if a+b and ab are alg/K then K(ab, a+b)/K is finite and now K(a, b)/K(ab, a+b) is at most a degree 2 extension, so K(a, b)/K is still finite, hence a, b, are alg/K

[somber sleet]

does it always hold that K(a+b) is the same as K(a,b)?

[rustic crown]

nope

b = -a

[somber sleet]

but it's contained

[rustic crown]

yee

[somber sleet]

cause I showed that K(a,b)/K is finite

does this automatically imply that K(a+b)/K also has to be?

[rustic crown]

yep

[somber sleet]

I think I'm missing a lil point here

[rustic crown]

its an intermediate field

[somber sleet]

Zwischenkörper for the win

thank youuuu

[rustic crown]

me watched one piece

[somber sleet]

getting a social pressure for you to watch one piece

[rustic crown]

hehe

now need to wait another week

[novel parrot]

i want to show this

if we take ideal in A/I

all ideals contain 0

and 0 in A/I corressponds to I in A

so any ideal in A/I contains atleast I (every ideal is atleast the 0 ideal)

does this work for converse

[coral shale]

Not sure, but I would advise sticking to doing it properly with injective/surjective first

before trying other stuff

[novel parrot]

okay

do you know what happens to the ideals that dont contain I ?

what happens to them in A/I

[coral shale]

I barely understand this problem in my head, just vaguely recall its an isomorphism theorem

I'd have to think of a concrete example to be sure.

iirc this is completely analogous with groups too

===
I feel like an ideal contains I only if it is a union of of cosets of I

So if it doesn't, that map produces something that isn't a coset

[oblique river]

What do you mean? You can consiser the image of them under the quotient map

And youll get some ideal of A/I

[novel parrot]

but ideals that dont contain I

intersect with only 0

[oblique river]

The image under the quotient map is some ideal of A/I

If you call tbe original ideal J, the image of J is the samw as the image of J+I, which is an ideal of A that contains I

[coral shale]

They mean consider this map on all ideals, not just the ones containing I J -> Jbar := {x + I : x in J}

right

and what it outputs

[novel parrot]

yes

so ideal J such that $I \cap J = {0}$

[cloud walrus]

ActiveChapter

[novel parrot]

the set of outputs is not an ideal anymore in A/I i think ?

[coral shale]

It just wont be a coset, I think.

Try with a concrete example and see what happens

[novel parrot]

i think that too

[coral shale]

say Z

I = (2)

J = (3)

oh wait u want completely empty intersection

uh

Think of a better example

[novel parrot]

i mean it doesnt matter just as long as J doesnt completely contain I

[coral shale]

alright then try that and see what happens

J -> Jbar := {x + (2) : x in J}

[novel parrot]

yeah

[coral shale]

looks like u end up with the entire quotient ring

[novel parrot]

maybe in more complicated rings it may not remain an ideal anymore?

[coral shale]

im not thinking particularly clearly rn

[oblique river]

No

The image of an ideal under any surjective map

Is still an ideal

It doesnt matter how it intersects the kernel of the map

[rustic crown]

Yui

[coral shale]

det is here to do the thinking, good.

[oblique river]

Am i invisible 😭

[coral shale]

certainly not, but hes a direct upgrade on me

[rustic crown]

send yui emote

but big

[coral shale]

Someone has to hold the weeb fort

[novel parrot]

i dont understand

[rustic crown]

okie what was the question uwu

[novel parrot]

ideals of A containing I are in bijection with ideals of A/I

what about ideals in A not completely containing I

[oblique river]

Theyre just not part of the correspondence

Again: the image of an ideal under a surjective map is still an ideal

Proof: just do it

[novel parrot]

okay so ideals get deleted when sending to A/I ?

"deleted"

[oblique river]

What do you mean

Multiple ideals of A can become the same in A/I

[novel parrot]

thats what im ask

it makes sense then

[oblique river]

Again: if J is any ideal of A, then the image of J is the same as the image of J + I

And if J containsn’t I, then J + I and J are different

[rustic crown]

containsn't

cute word

[oblique river]

Ty i just came up with it :)

[rustic crown]

i told a friend that i learned a new word. but i was saying it zwishenkorper, with z like a z and not s

.<

he told me i should know that much after staying 4 months in germany

[somber sleet]

OH

So you stayed 4 months in Germany

come on as long as you don't have to say Eichhörnchen everything is savable

[somber sleet]

I don't see what's wrong though

[rustic crown]

yee me doing masters here

[rustic crown]

[rustic crown]

he thought it sounded cute

[pastel cliff]

is this enough proof

[south patrol]

No

Because you didn't check addition

Though that is just as easy

[formal ermine]

what does ann mean? I've only seen it as tor

[south patrol]

And I would add smth at the end saying "hence ..."

Ann is defined there

The annihilator of a module

[formal ermine]

ah annihilator

[south patrol]

though unfortunately different to annihilators in lin alg ig

[formal ermine]

oh tor is like the opposite thing

oops

[south patrol]

Huh

[pastel cliff]

you mean the underlying abelian group right?

[south patrol]

Oh do you mean torsion submodule

I assumed you meant the tor functors lol

[south patrol]

Uhh I mean like

Ideals are closed under addition

You need to show that that is the case

[pastel cliff]

[south patrol]

Sorry yes there is a fancier way

But it is the same proof basically lol

You can think of this as the kernel of a map out of R

[toxic zephyr]

i believe you can do both at once by showing for r1,r2 in AnnR(M) and s1,s2 in R, then s1r1+r2s2 is in AnnR(M)

then s1=s2=1 shows closed under addition
s2=0 shows left ideal property
s1=0 shows right ideal property

for a comm ring you only nead sr1+r2

in the words of my professor, "you'd probably confuse a class teaching it like that, but it does work"

[astral stream]

What do these words mean, individually I understand but when they bunch them together my brain rots

@.me when replying so ik

[rotund aurora]

permutation multiplication is ordinary composition of functions

[ruby sundial]

@astral stream A collection of permutations are a collection of bijections f:A->A.

So think if this $S_A={f:A->A|$ f is a bijection $}$ and composition of these $f,g\in S_A$ is normal function composition. Composition of bijections is a bijection so the operation is closed. Now think about associativity,inverses, and identity.

[cloud walrus]

small saint

[chilly ocean]

\to

[ruby sundial]

lol i forgot latexing on phone is different

[astral stream]

oh wow thanks guys

that makes a lot of sense now

[ruby sundial]

what was your misunderstanding originally?

so that i can improve explanation

[astral stream]

For 'collection of all permutation of A' I didn't understand 'collection' = set and permutations referred to the function and not just a new set (so I thought permutation of {1,2} would just be {1,2} and {2,1} and not the functions). not sure if that makes sense

For 'group under permutation multiplication' I forgot permutation multiplication = function composition.

[toxic zephyr]

so this phi is not necessarily a group homomorphism, right? then would showing ker(phi)={1} not prove it is injective?
the second part of the question supposes H and K are finite and uses the cardinality to show that |HK|=|H||K|/|H∪K|. but i'm also not really understanding how we can go about showing |HK/K|=|HK|/|K|.

the notation HK/K is very suggestive. but i don't understand why {hK:h in H} would have the same cardinality as |HK|/|K| (when they're finite)

[oblique river]

Just by counting. Each coset of K in HK has the same size, and they are all disjoint

And their union is all of HK

[rotund aurora]

You can always decompose any finite separable extension as a tower of extensions of prime degree, right? I guess this is group theory

so you have sylow groups

and so you can reduce to the case where the extension has degree p^n

and you want to decompose it as n extensions of degree p

which I think is enough by Cauchy's theorem

no?

[oblique river]

What if it’s not galois?

[rotund aurora]

ah sorry forgot to include that hypothesis

I was thinking about Galois extensions, yeah

if its not Galois then idk, but I guess its still possible?

[oblique river]

I know examples of degree 4 extensions with no subfields

[rotund aurora]

nice

[oblique river]

But not galois

[rotund aurora]

yeah, what I said is true for Galois no? I made the sketch of the group theory

[toxic zephyr]

right, HK is the union of all left h cosets of K. so to count the number of cosets, we just divide the total number of elements by the size of each coset? just want to make sure i understand

[oblique river]

Even for galois fields, subfields need not be galois

So like take your first sylow, you get a subextension whose degree has no powers of p in it, but it might not be galois; how do you continue?

[oblique river]

Thats right

[coral spindle]

It is exactly the same as the typical proof of Lagrange

[rotund aurora]

you want everything to be as Galois as possible always, maybe taking the Galois closure takes us somewhere?

[toxic zephyr]

so can this application of Lagrange's theorem always apply to quotients, even ones that aren't groups?

[rotund aurora]

uhm wait no that doesnt make sense

wait Im confused about the fundamental theorem

[oblique river]

Like, take an A_5 extension. Im not seeing how your argument works.

[coral spindle]

Where the equivalence classes of are all the same size, yes

[toxic zephyr]

very cool. thank you both ❤️ @coral spindle @oblique river

[oblique river]

Nvm A_5 is fine, because it has A_4 as a subgroup

[rotund aurora]

Consider the extension $K<L$, let $G=\text{Gal}(L/K)$. Suppose $|G|=p_1^{a_1}\cdots p_n^{a_n}$. Let $H$ be a subgroup of $G$ of order $p_1^{a_1}$ (Sylow) and let $L^H$ be the fixed field. We have the tower $K<L^H<L$ where $[L^H\colon K]=p_2^{a_2}\cdots p_n^{a_n}$ and $[L\colon L^H]=p_1^{a_n}$

[cloud walrus]

Croqueta

[rotund aurora]

and $L^H<L$ is Galois

[cloud walrus]

Croqueta

[rotund aurora]

maybe Im making everything up lol, I have to check the fundamental theorem all the time

is that correct?

ah but you dont know if K<L^H is Galois

is that the problem?

[oblique river]

Yes

[rotund aurora]

So given a group G you would want to find subgroups G1<G2<...<Gn such that [G{i+1} : G{i}] is prime

this is the group theoretic formulation of what I was saying (in a Galois extension), correct?

that seems pretty impossible actually, idk

[south patrol]

Indeed this is just saying G is solvable

[rotund aurora]

true

lmao

because the quotients are of prime order, therefore super abelian

[south patrol]

Okay tbf I was sloppy there

If the series is subnormal this is solvabulity, but I assume you want normality cause we want galois extensions0

[rotund aurora]

ah right, no assumption on normality

I dont see why would you need normality at all

subfields correspond to subgroups, and the degree and index is preserved

[oblique river]

This isnt quite solvability

You can do this with A_5 for example

[rotund aurora]

I think what I was saying asks for the following: Given a group G, find a subgroup H of G such that [G : H] is a prime number.

Let K<L be an extension G=Gal(L/K). Say H is a subgroup of G with [G : H] prime. Then K<L^H<L with K<L^H of prime degree, L^H<L Galois and Gal(L/L^H)=H, so you can keep climbing

but I guess its just not true in general

[pastel cliff]

https://www.youtube.com/watch?v=IvukAijXgLE

these explainers are surprisingly good

[rustic crown]

role-module

[frank cosmos]

Lang left this as an exercise but is it just obvious? "The normalizer of a subgroup H of G is the largest subgroup of G where H is normal." Like if there is a bigger subgroup, all elements of it would be in the normalizer?

[next obsidian]

Yup

Basically by definition of the normalizer right?

[tawdry apex]

Hi! I'm supposed to factor (x^4+1) in the finite fields (\mathbb{F}5), (\mathbb{F}{5^2}) and (\mathbb{F}_{5^3}). And while this was easily done through elementary methods for the first case, I suspect I'm supposed to be able to use that knowledge as I proceed, but I'm just at a loss for how I'm meant to continue

[cloud walrus]

Marci

[next obsidian]

What did you get for F_5?

@tawdry apex

[rustic crown]

[next obsidian]

ShiN, back off

:(

I got this

[tawdry apex]

I'm starting to think I don't get how finite fields work but doesn't (x^2+2)(x^2+3) do the job?

[next obsidian]

Right, so let’s test it

If you naively expand it you get what?

[tawdry apex]

x^4+5x^2+6 which reduces to x^4+1 mod 5 right?

[next obsidian]

Yup!

Now does either of those factor even further?

[tawdry apex]

Just based on plugging in every element in each case I'm inclined to say no? (They're quadratics and have no roots)

[next obsidian]

Yes!

So the next part depends on how much you know about finite fields

1: do you know that F_5 < F_5^2?

< meaning “is a sub field”

[tawdry apex]

Yep

[next obsidian]

Cool, so first general fact

If K < L, and you have a polynomial in K[x] which factors in K, the same factorization holds over L

[next obsidian]

Does that make sense? What I’m saying here is that at the very least x^4 + 1 factors like this over F_5^2

Cool, so now we have to ask, does it factor even more now?

[tawdry apex]

Yep this is the point I got up to basically

[next obsidian]

Ah oky cool.

[tawdry apex]

I'm not sure if it does/ if we have any standards methods to see how

[next obsidian]

So do you know how to write F_5^2 as an extension over F_5?

Or I guess actually let’s be more general

What facts do you know about finite fields?

I’ll be even more pointed to help out

What even is F_5^2?

[tawdry apex]

OH! I know F_5^2 is the splitting field of x^(5^2)-x over F_5? Could that help?

[next obsidian]

Yes this may be helpful, but there’s something that might be even more helpful

How many finite fields exist

Of any given order?

[tawdry apex]

I thought they were unique?

[next obsidian]

!!!

Yes they are!

So now we’re in business

[next obsidian]

So we want to know if these two polynomials factor in F_5^2 yeah?

[tawdry apex]

(quick sidenote, thank you so much you're top tier at walking me through this I am beyond thankful already)

[next obsidian]

So let’s look just at x^2 + 2 first yeah?

This factors somewhere right?

[tawdry apex]

Yeah I guess if we just extend the field with one of its roots

[next obsidian]

Yeah exactly

What’s the order of that field

[tawdry apex]

OH it's a degree 2 extension so that extension has to be exactly F_25???

[next obsidian]

:D

What about the other polynomial?

[tawdry apex]

Ah that's quite cool how the uniqueness was important

That's also degree 2 so does that mean we can follow the same argument and both of them have to factor in F_25?

[next obsidian]

:D

But now we have to deal with F_125 right?

[tawdry apex]

Well it can't factor further than linear factors so it'll just be the same as for F_25

[next obsidian]

Well no!

[tawdry apex]

Damn

[next obsidian]

Does F_125 contain F_25?

[tawdry apex]

Ohhh

No it does not

[next obsidian]

So could those quadratics factor in it?

[tawdry apex]

Nope

[next obsidian]

:D

[tawdry apex]

I suddenly see why this problem was written in the first place

Wow thank you so much 😭 fields have never made more sense

Do you have any general tips on how to lay out similar hands on questions with more clarity? I often get stuck just not knowing how I'm supposed to even apply my knowledge

[next obsidian]

Idk

Experience

¯_(ツ)_/¯

I just rummage through a textbook trying to find relevant theorems

[tawdry apex]

Fair enough thank you for your help anyway

[long nebula]

projective resolutions are confusing me so much

pain

[next obsidian]

Haha noob

Tell Chmonkey your woes young padawan

[long nebula]

So like... since the sequence splits, each module is like isomorphic to the direct sum of the ones next to it, so then like P_1 is P_0 + P_2, but P_2 is itself P_1 + P_3, so P_1 is P_0 + P_1 + P_3 and that hurts my head

[rustic crown]

i don't understand 2 out of the 6 words

[next obsidian]

No it doesn’t

[long nebula]

Oh wait

[next obsidian]

You only surject onto the kernel of the last map

There’s no reason that’s projective

[long nebula]

The splitting thing only applies to a short exact sequence

Okay that's why I was confused

Lol nvm

Oopsie moment

[next obsidian]

.

:)

[rustic crown]

[next obsidian]

Damn I thought that sas Walter

[long nebula]

Okay this makes a lot more sense now lol

[rustic crown]

(will you not enlighten det? )

[next obsidian]

Oh MY sentence?

Lmfaooo

[long nebula]

Lol which two words

[next obsidian]

Padawan = apprentice Jedi, young padawan is just a meme term for like, a beginner

[rustic crown]

woes padawan

[next obsidian]

Don’t use it

Woe = worries basically

[rustic crown]

oooooh

[long nebula]

I thought you meant projective resolution for a sec

[rustic crown]

tell chmonkey your worries beginner

[long nebula]

I was getting ready to explain lol

[rustic crown]

hehe

[rustic crown]

[next obsidian]

It’s a starwars reference, it’s pretty cringe to say it lol

But I did it on purpose sarcastically so it’s marginally better

[long nebula]

so would an example just be
[\begin{tikzcd}\dots\arrow[r]&F(\ker d_1)\arrow[r,"d_2"]&F(\ker d_0)\arrow[r,"d_1"]&F(A)\arrow[r,"d_0"]&A\arrow[r]&0\end{tikzcd}]

[rustic crown]

yee

[cloud walrus]

tietzeric extension (he)

[long nebula]

ayyyy

okay

this is making sense now 🥺 I was overthinking it

[rustic crown]

🫂

Walterrrr

[agile burrow]

hi det

[rustic crown]

uwu

[agile burrow]

hope you're well and that you had a good weekend

[rustic crown]

(that's a nice functorial free resolution :3)

[long nebula]

:o words

[rustic crown]

me binge-re-watched vinland saga uwu

[rustic crown]

all it means is if A --> B is a map, then you get a ladder of the resolutions above

[agile burrow]

sounds like a weekend well spent

[long nebula]

So I guess I should ask next

Why are these useful

[rustic crown]

so it's like you want to study the object A

which is complicated

but free modules are not complicated

so you try to approximate A using free modules

think of generators and relations idea

[long nebula]

oooh

[rustic crown]

for PIDs these terminate (and f.g. modules)

but for general rings you might have to go on forever

[coral spindle]

They're the same as free resultions, up to homotopy

[long nebula]

Wait what does that mean

[coral spindle]

and free resolutions are much easier to motivate; they're generalised presentations

[long nebula]

Where does the homotopy come from

[coral spindle]

chain homotopy

If you've seen simplicial homology, it's like the prism map

[long nebula]

I haven't seen any of these

[coral spindle]

or whatever it's called. I think it's called the prism map.

OK well you'll get there.

There's a notion of homotopy.

[rustic crown]

so in some sense the object A
and the whole sequence
... --> F2 --> F1 --> F0 --> 0 --> 0 -->
have the same information

[long nebula]

Coool

[coral spindle]

I'm surprised you're looking at any homological algebra without first knowing simplicial homology lol

[rustic crown]

one entirely consists of free stuff which hopefully are easier to understand uwu

[coral spindle]

like how do you expect to motivate it

[long nebula]

It's just an algebra hw assignment

Lol

It wasn't anything complicated, it was just prove that any module has a projective resolution

But it seemed interesting

We just learned about modules and projective modules so that's why

Next class is on injective modules and baer's criterion 👀

So yeah the motivation might be over my head

[rustic crown]

you know cat theory right

[long nebula]

Very little

[rustic crown]

okie that's enough :3

[long nebula]

I've been reading the first chapter of riehl's category theory in context

[rustic crown]

so you get a functor
R-mod --> Ch(R-mod)

[long nebula]

what's Ch(R-mod)?

[coral spindle]

Grrr that's a lie

[long nebula]

Chmonkey(R-mod)?

[rustic crown]

yes

:3

[coral spindle]

:( should be a functor up to homotopy alone

[long nebula]

(R-mod)

[rustic crown]

Ch(R-mod) are chain complex for R-moduels

[long nebula]

Okay next question

What is a chain complex

😭

[rustic crown]

stuff you have been looking at

[long nebula]

Okay

[rustic crown]

--> M2 --> M1 --> M0 --> M-1 --> ...

[long nebula]

So the projective resolution is an example of that?

[rustic crown]

where composite of two adjacent maps is 0

[long nebula]

Oh okay

So instead of exactness, the image is just included in the kernel

[rustic crown]

so the functor above sends a module M to the trivial complex