#groups-rings-fields | Mathematics | Page 465

ripe crest Mar 17, 2020, 4:38 AM

#

yup

oblique river Mar 17, 2020, 4:38 AM

#

but now, in the quotient, we can replace y with x^2 + x + 1

#

so our ideal is (x-a, (x^2+x+1) - (a^2 + a + 1))

#

= (x-a, (x^2 - a^2) + (x-a))

#

= (x-a, (x-a)*(x + a + 1))

#

= (x-a)

#

basically when we quotient by "y = blah" it means we can basically forget abotu y

#

since we're writing it in terms of x

#

it's like in group theory if I tell you that a group is generated by a and b with the relation b = a^2

#

then we really don't need to talk about b any more

ripe crest Mar 17, 2020, 4:40 AM

#

ah, I see

#

Hmmm I feel like I'm missing something

#

You're saying in the quotient, that we're just left with ideals generated by a single polynomial

#

Ah

#

And instead of just saying that it's isomorphic to C[T], you showed me by hand why it's true

#

I get it

oblique river Mar 17, 2020, 4:45 AM

#

👍

ripe crest Mar 17, 2020, 4:45 AM

#

lmao i had to figure out what was even happening

#

Okay, then going back to finding prime ideals

#

Prime ideals of the quotient are a subset of the ideals in the quotient, which themselves are (images of) maximal ideals of C[X,Y]

#

oh wait

#

that's not true xD

#

buuut! Prime <=> maximal in the quotient. So, the prime ideals are precisely the maximal ideals!

#

ayo let's go

oblique river Mar 17, 2020, 4:48 AM

#

haha

#

you know that prime = maximal in the quotient because it's isomorphic to C[t]

#

the other way to think about it is that prime ideals in the quotient are exactly the images of prime ideals upstairs

#

well actually taht's hard to think about since there are lots of those

#

that aren't maximal

#

yeah probably saying "iso to C[t] so prime = maximal" is the best way to go

ripe crest Mar 17, 2020, 4:49 AM

#

oh? if we have a surjective homomorphism then maximal -> maximal, and prime -> prime?

oblique river Mar 17, 2020, 4:50 AM

#

yes that is the correspondence theorem

#

or sometimes given a "Nth isomorphism theorem" name

#

for some N

#

there is an inclusion-preserving bijection between {ideals of R containing I} and {ideals of R/I}

#

and that correspondence preserves maximality and primality

ripe crest Mar 17, 2020, 4:51 AM

#

interesting. I knew preimages of ideals being ideals and images of ideals being ideals when the map is surjective - but didn't know about that

#

I guess it just follows from that

#

at least the maximality part does for the quotient ring

oblique river Mar 17, 2020, 4:52 AM

#

if J contains I in R, the image of J in R/I is J/I

ripe crest Mar 17, 2020, 4:52 AM

#

right

oblique river Mar 17, 2020, 4:52 AM

#

and some other isomorphism theorem tells you that R/J = (R/I)/(J/I)

#

so if one of those rings is an integral domain/field, the other must be as well

solemn rain Mar 17, 2020, 4:52 AM

#

^ 3rd

ripe crest Mar 17, 2020, 4:53 AM

#

gotcha

solemn rain Mar 17, 2020, 4:53 AM

#

there is an inclusion-preserving bijection between {ideals of R containing I} and {ideals of R/I}

#

^ fourth

ripe crest Mar 17, 2020, 4:53 AM

#

and no one cares about the 2nd xD

solemn rain Mar 17, 2020, 4:53 AM

#

yea thank god

oblique river Mar 17, 2020, 4:53 AM

#

also don't forget that if R --> S is surjective then S is of the form R/I

#

so this is all true for any surjective map

ripe crest Mar 17, 2020, 4:53 AM

#

ah, by the 1st

oblique river Mar 17, 2020, 4:53 AM

#

it's even true for nonsurjective maps that the inverse image of a prime ideal is prime

#

but there the inverse image of a maximal ideal need not be maximal

#

proof: Let p be a prime of S and P the inverse image in R (for some hom R --> S)

#

then P is the kernel of the composite R --> S --> S/p

#

by 1st iso thm, this means that hte image of that map is iso to R/P

#

i.e. R/P is a subring of S/p

#

if S/p is an integral domain, then R/P must be as well

#

R/P might not be all of S/p since the original map wasn't assumed to eb surjective

#

we also see why it fails for maximal ideals: subrings of fields need not be fields

ripe crest Mar 17, 2020, 4:57 AM

#

oh that's nice

#

R/P injects into S/p

#

that's actually quite clever

#

that composite map

#

I've seen that before, guess it's a useful technique

oblique river Mar 17, 2020, 4:58 AM

#

yep

#

it's the same way you prove the third iso theorem

#

consider the composite R --> R/I --> (R/I)/(J/I)

#

this is surjective as both pieces are

#

and the kernel is J

#

hence R/J is iso to (R/I)/(J/I)

ripe crest Mar 17, 2020, 5:01 AM

#

How do we know kernel is J?

#

I guess we because J contains I first off, and then we mod out by J

oblique river Mar 17, 2020, 5:01 AM

#

the kernel is the things that map to 0 in (R/I)/(J/I)

#

i.e. the things that map to J/I in R/I

#

and yeah since J contains I

ripe crest Mar 17, 2020, 5:02 AM

#

gotcha, cool stuff

#

are you in grad school?

oblique river Mar 17, 2020, 5:05 AM

#

yeah

#

I'm finishing this spring

ripe crest Mar 17, 2020, 5:05 AM

#

oh wow

#

nice

#

in some part of algebra? Algebraic goemetry?

oblique river Mar 17, 2020, 5:06 AM

#

I do number theory

#

and thanks :)

ripe crest Mar 17, 2020, 5:06 AM

#

elliptic curve stuff?

oblique river Mar 17, 2020, 5:08 AM

#

I think about elliptic curves sometimes but it's not my specialty

#

I'm interested in class groups of number fields

#

and galois cohomology

ripe crest Mar 17, 2020, 5:10 AM

#

sounds very cool - didn't really understand it on wikipedia 😛

oblique river Mar 17, 2020, 5:10 AM

#

haha

ripe crest Mar 17, 2020, 5:10 AM

#

Are you planning to stay in academia?

oblique river Mar 17, 2020, 5:10 AM

#

I'm a little preoccupied right now but I'd be happy to give you a synopsis at some point

#

yeah though not in a pure research role

#

I'm taking a teaching-focused position next year

ripe crest Mar 17, 2020, 5:11 AM

#

Preferably after my midterm : P

#

oooh very nice

#

In America?

#

It's fine if you don't want to say

oblique river Mar 17, 2020, 5:12 AM

#

yeah

#

in st louis

#

I'm currently in chicago

ripe crest Mar 17, 2020, 5:12 AM

#

goddamn

#

chicago math is known to be solid

#

all I know about st louis are the blues (hockey team)

oblique river Mar 17, 2020, 5:13 AM

#

haha

#

I'm not much of a hockey person :P

ripe crest Mar 17, 2020, 5:14 AM

#

me neither, used to be tho

#

I am canadian though

oblique river Mar 17, 2020, 5:14 AM

#

ah gotcha

ripe crest Mar 17, 2020, 5:16 AM

#

Here's a question: Find an idempotent element of R[X]/(X-2)(X-1) which is not 0 or 1

#

and apparently there's a way to relate this to the chinese remaiender theorem

#

But i dont know that, but i need to know it for tomorrow

oblique river Mar 17, 2020, 5:17 AM

#

what does CRT say about R[x]/(x-2)(x-1)

ripe crest Mar 17, 2020, 5:17 AM

#

no idea - lemme look

#

R[X]/(X-2)(X-1) is isomorphic to R[X]/(X-1) x R[X]/(X-2)

#

oh cool

#

that's just R x R, which is R^2

#

and maybe the multiplication behaves like C

#

so we want z \in C s.t. z^2 = z

#

so it has to be on unit circle

oblique river Mar 17, 2020, 5:21 AM

#

careful

#

R x R is not iso to C

#

as rings

ripe crest Mar 17, 2020, 5:21 AM

#

😦 if only

oblique river Mar 17, 2020, 5:21 AM

#

in R x R, (a,b)*(c,d) = (ac,bd)

#

which is not how multiplication works in C

ripe crest Mar 17, 2020, 5:21 AM

#

oh yeah duh wtf

oblique river Mar 17, 2020, 5:21 AM

#

;)

ripe crest Mar 17, 2020, 5:22 AM

#

ok, there goes that plan of attack

#

still, we want (a, b) s.t. (aa, bb) = (a, b)

#

But, like R is an integral domain - so either a = 0 or a = 1..

#

hmm

oblique river Mar 17, 2020, 5:23 AM

#

correct

#

what are the elements "1" and "0" in RxR?

ripe crest Mar 17, 2020, 5:24 AM

#

1 would be (1, 1) and 0 would be (0, 0)

#

Oh

#

(1, 0), (0, 1)

#

those are 2 candidates for idempotents

oblique river Mar 17, 2020, 5:25 AM

#

yep

ripe crest Mar 17, 2020, 5:25 AM

#

that are not 0 and not 1

#

Let me see..

#

What would (1, 0) look like in the original quotient ring

#

It's 1 under quotienting out by (X-1), and 0 quotienting out by (X-2)

#

hmm I don't know

#

guessing and testing isn't working

#

ah well - finding things up to an isomorphism should be good enough on the midterm

oblique river Mar 17, 2020, 5:31 AM

#

what about x-2?

ripe crest Mar 17, 2020, 5:31 AM

#

😦

#

wait

#

no - that won't work

#

because it'll be 0 in the original quotient

oblique river Mar 17, 2020, 5:32 AM

#

oop you're right

#

sry

#

so here's another way to think about it

#

we need to find a poly such that when you plug in 2 you get 0

#

and when you plug in 1 you get 1

#

so it has to be a multiple of x-2

#

and in fact we can see that if f(x) = x-2 then f(1) = -1

#

so choose -x + 2

#

f(2) = 0 and f(1) = 1

#

-x+2 is 0 mod x-2 and 1 mod x-1

#

we can also check it's an idempotent directly (although it's unnecessary -- we just proved it's an idempotent, but it's always nice to double check). (2-x)^2 = x^2 - 4x + 4.

#

we are modding by (x-2)(x-1) = x^2 - 3x + 2

ripe crest Mar 17, 2020, 5:35 AM

#

wait - you were totally right before! x-2 does work!

#

because if -x+2 works then x-2 must work

oblique river Mar 17, 2020, 5:35 AM

#

(x^2 - 4x + 4) - (x^2 - 3x + 2) = -x + 2

ripe crest Mar 17, 2020, 5:35 AM

#

Because we factor by (x^2 -3x +2)

#

and x-2 -> x-2

oblique river Mar 17, 2020, 5:35 AM

#

x-2 won't work here. x-2 in the RxR coordinates is (-1,0)

#

so its square is (1,0)

#

not itself

ripe crest Mar 17, 2020, 5:36 AM

#

ah ofc

#

that's really nice

#

You just chose Q in P = (X-2)Q such that P(1) would equal 1

#

so obvious in hindsight

#

yet

#

so far away from my grasp

oblique river Mar 17, 2020, 5:38 AM

#

that's what years of experience will do for you :)

ripe crest Mar 17, 2020, 5:44 AM

#

A is a commutative ring, e an idempotent of A. Need to show that A is isomorphic to Ae x A(1-e)

#

gonna spend some time with this

#

don't spoil it yet xD

oblique river Mar 17, 2020, 5:45 AM

#

ok :)

ripe crest Mar 17, 2020, 5:49 AM

#

Aside - a homomorphism of k-algebras in the context of single-variable polynomials rings is completely determined by where it sends X right?

#

and if it was multivariate, where it sends X_1, X_2, ..., X_n

mild laurel Mar 17, 2020, 5:50 AM

#

yes

smoky cypress Mar 17, 2020, 5:54 AM

#

Let G be a cyclic group of order n and r|n. How do I prove that there is only one subgroup H of G with |H|=r?

solemn rain Mar 17, 2020, 5:54 AM

#

fundamental theorem

smoky cypress Mar 17, 2020, 5:55 AM

#

can you elaborate on that?

solemn rain Mar 17, 2020, 5:55 AM

#

suppose another 111

#

oka

#

okay lets do it togethere

#

let me state it this way :

#

let G = <x> be a cyclic group

#

with order n.

#

for each divisor of n,m, there exists a unique subgroup of G

#

namely <x^(n/m)>

#

we want to prove this

smoky cypress Mar 17, 2020, 5:56 AM

#

yes

solemn rain Mar 17, 2020, 5:57 AM

#

now let k = n/m

#

can u show that <x^k> is a subgroup?

#

then we show uniqunees together

smoky cypress Mar 17, 2020, 5:57 AM

#

It is

solemn rain Mar 17, 2020, 5:57 AM

#

okay so u struggle with unique?

smoky cypress Mar 17, 2020, 5:57 AM

#

Yeah

solemn rain Mar 17, 2020, 5:58 AM

#

okay so lets call this subgroup H

smoky cypress Mar 17, 2020, 5:58 AM

#

I proved that subgroup of cyclic group is cyclic

solemn rain Mar 17, 2020, 5:58 AM

#

so suppose there exists another subgroup G' with order k

smoky cypress Mar 17, 2020, 5:58 AM

#

Ok

#

So H=<x^r> and G'=<x^t>

solemn rain Mar 17, 2020, 5:59 AM

#

yea where r and t are the samllest possible intgers where x^n is in H and x^t is in G' respectively

#

thats from what you proved

smoky cypress Mar 17, 2020, 5:59 AM

#

x^r in H

#

yes

solemn rain Mar 17, 2020, 5:59 AM

#

oh my god i g ot lost up

#

with the names

#

okay lets summarize sorry

#

let H = <x> be a cyclic group with finite order n

#

for every divisor ,a, there is aunique subgroup namely <x^(n/a)>

#

now proof:

#

assume H = <x> with finite order n

#

and a is a divisor of n

#

now suppose K is any subgroup of H of order a. ( to show uniqueness )

#

then we have K = <x^b> where b is the smallest positive integer such that x^b is in K

smoky cypress Mar 17, 2020, 6:02 AM

#

I think got it now xD

solemn rain Mar 17, 2020, 6:02 AM

#

really?

#

anyways

#

|K| = |x^b| = n/(n,b)

smoky cypress Mar 17, 2020, 6:03 AM

#

so b is the smallest positive integer with x^b in K

#

and since K has order a, we have that x^{ab}=1, so n|ab

#

or n<=ab, and b>=n/a

#

and x^{an/a}=x^n=1 shows that b<=n/a

#

therefore b=n/a, the obvious choice

solemn rain Mar 17, 2020, 6:04 AM

#

okayy

#

well i did nothing to help ig XD

#

u figured out on ur own

#

sorry for the confusion earlier

ripe crest Mar 17, 2020, 6:04 AM

#

That's the best help!

smoky cypress Mar 17, 2020, 6:04 AM

#

Lmao

#

That's fine

solemn rain Mar 17, 2020, 6:05 AM

#

if u have any problems post here

#

it would help me too as im practicing

#

this is called the fundamental theorem of cyclic groups

#

3 points actually

#

every subgroup of cyclic groups is cyclic

#

order of cyclic subgroup must divide group ( hehe xd )

#

and for every divisor of order of finite cyclic group there exists a unique cyclic subgroup of that order

#

now that

ripe crest Mar 17, 2020, 6:06 AM

#

@oblique river i have in my course notes that if I have orthogonal idempotents, I can decompose my ring as a product of them

solemn rain Mar 17, 2020, 6:07 AM

#

with |x^a| = |x|/(|x|,a)

#

u finished cyclic groups

#

basiocally

stone fulcrum Mar 17, 2020, 6:07 AM

#

Order of any subgroup divides the order of the group, that's not only cyclic groups

ripe crest Mar 17, 2020, 6:07 AM

#

Oh, I look at the ideals generated by them

solemn rain Mar 17, 2020, 6:08 AM

#

yea thats why i added the hehe xd part

smoky cypress Mar 17, 2020, 6:08 AM

#

Yeah that's why he said the (hehe xd)

#

lol

solemn rain Mar 17, 2020, 6:08 AM

#

lmao

stone fulcrum Mar 17, 2020, 6:08 AM

#

Fair I get it nao

solemn rain Mar 17, 2020, 6:08 AM

#

cuz i think this section was b4

ripe crest Mar 17, 2020, 6:08 AM

#

they are comaximal

solemn rain Mar 17, 2020, 6:08 AM

#

the lagranges theorem section

smoky cypress Mar 17, 2020, 6:08 AM

#

well

stone fulcrum Mar 17, 2020, 6:08 AM

#

Inb4 Lagrange

smoky cypress Mar 17, 2020, 6:08 AM

#

i wrote |x^a|=lcm(|x|,a)/a

#

xD

#

Same thing

solemn rain Mar 17, 2020, 6:08 AM

#

nah its gcd

ripe crest Mar 17, 2020, 6:08 AM

#

interesting that this means that (2, 3) generates Z

smoky cypress Mar 17, 2020, 6:09 AM

#

It's the same

solemn rain Mar 17, 2020, 6:09 AM

#

yea it hink thats why u use the notation (a,b) for gcd(a,b)

#

cuz for osmething in rings

#

something*

smoky cypress Mar 17, 2020, 6:09 AM

#

ideals

solemn rain Mar 17, 2020, 6:09 AM

#

yea idk

#

abotu them

ripe crest Mar 17, 2020, 6:09 AM

#

yeah

solemn rain Mar 17, 2020, 6:19 AM

#

I DONT UNDERSTAND COMPOSITION SERIES

#

they are important arent theyu

ripe crest Mar 17, 2020, 6:20 AM

#

oh btw @solemn rain - Rotman is really good for group theory - I liked his textbook

solemn rain Mar 17, 2020, 6:20 AM

#

whats rotman

#

advanced modern algebra?

ripe crest Mar 17, 2020, 6:20 AM

#

introduction to the theory of groups

#

https://www.amazon.ca/Introduction-Theory-Groups-Joseph-Rotman/dp/0387942858

solemn rain Mar 17, 2020, 6:22 AM

#

im currently using df

#

and it took alot of my time

#

i dont want to switch books tbh

#

i like that 1 tho

#

advnaced modern algebra looks coool too

ripe crest Mar 17, 2020, 6:23 AM

#

oh yeah for sure - i just meant as a supplement if you wanted to see a different way of doing things

#

sometimes a different author's explanations make more sense

smoky cypress Mar 17, 2020, 6:32 AM

#

📎 Screen_Shot_2020-03-16_at_11.32.01_PM.png

#

So this is at the end of Teacher's Notes in Artin's Algebra

#

What is he trying to say here?

#

That his text is not standard?

solemn rain Mar 17, 2020, 6:35 AM

#

maybe he is just trying to be cute

latent anvil Mar 17, 2020, 8:12 AM

#

Finished my algebra final

#

Can I get a "fuck Galois theory" in chat?

mild laurel Mar 17, 2020, 8:12 AM

#

tfw im looking at the wiki page for the primitive element theorem rn

latent anvil Mar 17, 2020, 8:13 AM

#

That's a mood

mild laurel Mar 17, 2020, 8:13 AM

#

you've gotta admit thats a nice theorem

latent anvil Mar 17, 2020, 8:13 AM

#

I was just looking at Hilbert theorem 90

#

It's a nice theorem

mild laurel Mar 17, 2020, 8:14 AM

#

galois theory has so many nice theorems

#

uh i mean

#

fuck galois theory amirite

tribal pasture Mar 17, 2020, 8:18 AM

#

So to show the isomorphism here, can I proceed like the following? Note that if f is R mod. hom. and g is R mod. hom. then (f,g) is R mod. hom.
Then we have that the following is an isomoprhism
[ \begin{tikzcd}
M \arrow[rightarrow]{r}{(\phi, g)}
& M' \oplus M''\arrow{r}{\varphi + f}
& M
\end{tikzcd}
]
If this is not true, can somebody guide on how to show the isomorphism, my prof just left it as "Prove the isomorphism"

📎 unknown.png

cloud walrusBOT Mar 17, 2020, 8:18 AM

#

All functions are continuous:

📎 626379040667992095.png

cloud walrusBOT Mar 17, 2020, 9:48 AM

#

Curiosity Zero:

📎 522733361031151616.png

lavish gale Mar 17, 2020, 9:49 AM

#

If we look at the contrapositive statement we get

cloud walrusBOT Mar 17, 2020, 9:50 AM

#

Curiosity Zero:

📎 522733361031151616.png

upper pivot Mar 17, 2020, 9:57 AM

#

the empty set satisfies the criteria

#

but isnt a subgroup

#

(but if you say non empty this will be sufficient)

harsh kettle Mar 17, 2020, 12:56 PM

#

can someone elucidate me on the importance of conjugacy across algebra , except for when you need exponentiation ?

chilly ocean Mar 17, 2020, 2:01 PM

#

i'm also interested in that question (if it's about groups)
the examples i've seen point to conjugates being analogous, with some renaming
so conjugate permutations have the same cycle structure but different names for the permuted elements
conjugate matrices are something similar with a change of basis
conjugation in quaternions is interesting, apparently it lets you rename e.g. i->j->k->i corresponding to a rotation

#

so an algebraist speaking chinese is conjugate to an algebraist speaking english, because if you put an english->chinese translator and a chinese->english translator (inverses) around the first one you get the second one

somber bramble Mar 17, 2020, 2:21 PM

#

but translations aren’t really invertible!

#

on the level of individual words anyway, translations are many-to-many relations

uncut girder Mar 17, 2020, 3:54 PM

#

Help I need help

#

The problem is as follows:

#

Let K be a finite extension of F. Prove that K is a splitting field over F iff every irreducible polynomial in F[x] that has a root in K spits completely in K[x].

wind steeple Mar 17, 2020, 5:06 PM

#

where are you struggling ?

uncut girder Mar 17, 2020, 5:15 PM

#

Forwards direction

#

I know K = F(a1, ..., an) where a1, ..., an are the roots of the polynomial for which K is the splitting field

wind steeple Mar 17, 2020, 5:24 PM

#

if you take Omega as an algebraic closure of F, every homomorphism from K to Omega is an automorphism because an homomorphism is determined by its value on a1,...,an.

#

the result follow almost directly

uncut girder Mar 17, 2020, 6:18 PM

#

Ah I see

solemn rain Mar 17, 2020, 6:48 PM

#

is the empty set subsuet of any set?

smoky cypress Mar 17, 2020, 6:49 PM

#

yes

#

vacuous truth

solemn rain Mar 17, 2020, 6:53 PM

#

t6y

chilly ocean Mar 17, 2020, 7:18 PM

#

Why are there two conjugacy classes of 11-cycles in A_12

#

i know the order of the centralizer of some 11-cycle σ in S_12 is 11, so centralizer is <σ>, which is contained entirely in A_12

#

tho idk why it follows that the single conjugacy class of σ in S_12 divides into two classes of equal size in A_12

chilly ocean Mar 17, 2020, 7:41 PM

#

is this channel occupied?

cloud walrusBOT Mar 17, 2020, 7:56 PM

#

creamy shits:

📎 210924359240384513.png

#

creamy shits:

📎 210924359240384513.png

mild laurel Mar 17, 2020, 7:59 PM

#

Uh, no? There are polynomials like f(x,y) = xy that are 0 whenever x is 0

latent anvil Mar 17, 2020, 8:03 PM

#

Let K/L be a Galois extension and G its Galois group. Are there any tricks for finding the representations of G?

#

Inspired by the problem "How many pairwise inequivalent irreducible complex representations does the Galois group (over Q) of the polynomial x^4+x^2+2x+1 have?" on my algebra final

#

One thing is noticed is that K is an L[G] module

bleak abyss Mar 17, 2020, 8:25 PM

#

Hmm, all the representations of G or just the ones involved in the Galois action?

#

I mean so

#

The Galois action can be done somewhat explicitly, since K is separable you can write it as L(\alpha)

#

So you have a basis {1,\alpha,...,\alpha^{d-1}}

#

And in principle you can write down the matrices in this basis, and use character stuff to decompose into irreps

latent anvil Mar 17, 2020, 8:27 PM

#

You're saying you can do this to the galois action of G on K?

#

That makes sense

bleak abyss Mar 17, 2020, 8:28 PM

#

Yeah

#

Now will this give all of them? Probably not

#

I wonder whether there's any other obvious info the Galois group contains. I could see something like that, e.g. the characters of the reps that do come up are algebraic (in fact cyclotomic) integers

#

So idk maybe you take the cyclotomic field generated by those guys and play some games

prisma skiff Mar 17, 2020, 9:46 PM

#

is it possible to prove closure from the other 3 properties of a group + abelian

#

given no information about the binary operator

mild laurel Mar 17, 2020, 9:46 PM

#

no, I mean

#

you can just consider the elements {-1,0,1} inside of Z under addition

prisma skiff Mar 17, 2020, 9:48 PM

#

mm i see, good example, thank you

solemn rain Mar 17, 2020, 9:55 PM

#

can some1 prove for me cauchys theorem

#

the induction one

chilly ocean Mar 17, 2020, 10:38 PM

#

?

solemn rain Mar 17, 2020, 10:48 PM

#

If G is a finite group and p is a prime dividing the order of the group

#

then there exists an element of order p

solemn rain Mar 17, 2020, 11:06 PM

#

and another problem too that i cant doo

#

do

#

definition : A subgroup of G ,H, is called a Hall subgroup iff (|H| , [G:H]) = 1

#

suppose H is a Hall subgroup of G and N is normal in G

#

prove that H intersects N is a Hall subgroup of N

chilly ocean Mar 17, 2020, 11:11 PM

#

@solemn rain For Cauchy's theorem, consider using the class equations.

solemn rain Mar 17, 2020, 11:11 PM

#

idk that yet

chilly ocean Mar 17, 2020, 11:12 PM

#

what are you allowedto use for now

#

have u covered group actions yet

solemn rain Mar 17, 2020, 11:12 PM

#

no thats next chapter

#

well

#

ik abit of them but like on a surface lvl

chilly ocean Mar 17, 2020, 11:13 PM

#

have you guys covered the orbit stabilizer theorem or something

solemn rain Mar 17, 2020, 11:13 PM

#

no

chilly ocean Mar 17, 2020, 11:13 PM

#

maybe in a specific setting rather than general

#

oh ok

fossil marsh Mar 17, 2020, 11:13 PM

#

The abelian case should be fairly simple without the use of either concept

chilly ocean Mar 17, 2020, 11:13 PM

#

he has to prove it generally

fossil marsh Mar 17, 2020, 11:14 PM

#

Hm

#

That is a bit tricky

chilly ocean Mar 17, 2020, 11:14 PM

#

The abelian case is indeed simple (in terms of machinery needed), but idk how to do half the proof if we cant use group actions or class equations

solemn rain Mar 17, 2020, 11:14 PM

#

can i use

#

okay

#

lets do abelian

#

then when i later cover group actions

#

i do for any group

chilly ocean Mar 17, 2020, 11:15 PM

#

Ok heres a hint for Abelian case

solemn rain Mar 17, 2020, 11:17 PM

#

for the hall problem

#

im given a more specific case that i can prove

#

maybe the generlization is harder

#

basically to prove intersection of any sylow p-subgroup of G with a normal subgroup N is a sylow-p subgroup of N

woven delta Mar 17, 2020, 11:18 PM

#

I don't think that's so hard

solemn rain Mar 17, 2020, 11:18 PM

#

okay will try that

chilly ocean Mar 17, 2020, 11:22 PM

#

Let n = |G|/p and consider when n=1. Clearly G is prime order cyclic group and hence contains an element of order p. Now suppose |G|=np and there is H with order kp, k<n. Continue from there

#

for cauchy's thm (for abelian grp)

solemn rain Mar 17, 2020, 11:23 PM

#

what is h

#

H*

chilly ocean Mar 17, 2020, 11:23 PM

#

oh sorry

#

H is a subgroup of G with order kp, where k<n

solemn rain Mar 17, 2020, 11:24 PM

#

okay will try

#

tysm

chilly ocean Mar 17, 2020, 11:24 PM

#

if you induct on the order of G then what can you say about H (by inductive hypothesis)

#

after you figure that out, then think about the orders of elements in these two groups (and consider what you can say about the quotient group G/H if you can construct one)

woven delta Mar 17, 2020, 11:27 PM

#

The proof is that you take a sylow p subgroup P of N. This is a p subgroup of G. Now we know that there is a sylow p subgroup P' of G st P subset of P'

#

Now we use the fact that P' and your original sylow p subgroup of G are congugate

#

And that N is normal

#

You can fill in the rest of the details yourself

#

@solemn rain

solemn rain Mar 17, 2020, 11:29 PM

#

yea ggot it

#

im trying this cauchy 1 first

#

tysm thoi

#

@chilly ocean G is abelian

#

H is normal

#

|G/H| = |G|/|H|

solemn rain Mar 18, 2020, 2:20 AM

#

i dont understand this definition already lmao

📎 unknown.png

mild laurel Mar 18, 2020, 2:22 AM

#

What don't you get?

solemn rain Mar 18, 2020, 2:24 AM

#

so is a composition series

#

a sequence of subgroups

#

?

#

such that subgroup mod the previous subgroup has no normal subgroups?

#

i dont get 1 = G tho

#

1 = N_0 <= N_2 ... = G?

mild laurel Mar 18, 2020, 2:33 AM

#

It just means that N_0 is the trivial group

#

And N_k is just G

solemn rain Mar 18, 2020, 2:38 AM

#

ohhhhhhhhhhhhhhh

#

yea got it

#

my bd

ripe crest Mar 18, 2020, 3:40 AM

#

heyo @oblique river

#

had my midterm

#

think i passed - didnt do great, but oh well

oblique river Mar 18, 2020, 3:41 AM

#

well passing is good :)

#

sorry it wasn't as good as you'd hoped

ripe crest Mar 18, 2020, 3:42 AM

#

There was a question where A is in M_n(k), where k is a field. And we consider the sub-k-algebra generated by A, call it B. And we were asked if it was a integral domain

#

it's fine - i could've studied more

#

couldn't fully get it. I looked at the map k[X] -> M_n(k), that sends P -> P(A)

#

so we look at k[A]

#

and if the kernel of that map was a prime ideal, then we were done

#

but I couldn't show that

#

and I left a couple others blank

oblique river Mar 18, 2020, 3:44 AM

#

the kernel depends on A

ripe crest Mar 18, 2020, 3:44 AM

#

Yeah

oblique river Mar 18, 2020, 3:44 AM

#

the kernel is generated by the [blank] of A

ripe crest Mar 18, 2020, 3:45 AM

#

hmmm

#

I know that the kernel is generated by a single element, since k[X] is a PID. I know that the kernel contains all polynomials s.t. P(A) = 0... i'm not sure what the blank is

oblique river Mar 18, 2020, 3:47 AM

#

what is the smallest polynomial such that p(A) = 0?

ripe crest Mar 18, 2020, 3:47 AM

#

the 0 polynomial

oblique river Mar 18, 2020, 3:47 AM

#

you're not wrong

ripe crest Mar 18, 2020, 3:47 AM

#

heh

#

i know that's not what you were asking 😛

oblique river Mar 18, 2020, 3:47 AM

#

what is the nonzero polynomial of minimal degree

#

the... minimal polynomial when ordered by degrees

ripe crest Mar 18, 2020, 3:49 AM

#

This must depend on A, no?

golden pasture Mar 18, 2020, 3:49 AM

#

||oao cauchy hamilton||

oblique river Mar 18, 2020, 3:49 AM

#

yes

ripe crest Mar 18, 2020, 3:49 AM

#

hmm, not looking at spoilers yet

oblique river Mar 18, 2020, 3:49 AM

#

it does depend on A

#

also if you didn't get my hint i'm not sure if you'll figure it out

#

the kernel is generated by the minimal polynomial of A

#

it's a polynomial that divides the characteristic polynomial

#

and tells you a lot about the matrix

ripe crest Mar 18, 2020, 3:50 AM

#

the determinant one?

oblique river Mar 18, 2020, 3:51 AM

#

what do you mean

#

the characteristic polynomial is a determinant

#

det(A-tI)

ripe crest Mar 18, 2020, 3:51 AM

#

right

#

ok im gonna look at the spoiler - my linear algebra might be lacking

#

no clue what cauchy hamilton is lol

oblique river Mar 18, 2020, 3:51 AM

#

I'm not sure if that theorem is really applicable here though

#

it says that a matrix A satisfies its characteristic polynomial

#

but it's not really useful here

ripe crest Mar 18, 2020, 3:52 AM

#

Okay, the kernel is generated by the minimal polynomial of A, whatever that means

oblique river Mar 18, 2020, 3:52 AM

#

here are some linear algebra facts:

#

the roots of the minimal polynomial are the eigenvalues of A

#

so if the quotient by the minpoly is going to be an integral domain, the minpoly can only have one root

#

and in fact the minpoly must be (t-a) for some a

#

this means that the matrix must be a*I

#

because that's the only matrix A which satisfies A - aI = 0

#

so it's an integral domain iff A is a scalar multiple of the identity

ripe crest Mar 18, 2020, 3:56 AM

#

yeah that makes sense

#

I never knew this stuff

#

hmm maybe we covered it in class, but I was too tired or not paying attention

mild laurel Mar 18, 2020, 3:57 AM

#

sorry, but does this depend on k being algebraically closed? otherwise can we know that the minimal polynomial splits into linear factors?

oblique river Mar 18, 2020, 3:57 AM

#

yeah, you're right

ripe crest Mar 18, 2020, 3:59 AM

#

hopefully i'll get 1/3 marks for at least getting started on the approach xD

#

are exact sequences cool/useful/important?

#

we're covering them now alongside modules

oblique river Mar 18, 2020, 3:59 AM

#

yes

#

basically just because theyc ome up everywhere

scarlet estuary Mar 18, 2020, 4:02 AM

#

i have typed $0 \to A \to B \to C \to 0$ so many times into $\LaTeX$ that i have a custom command for it

ripe crest Mar 18, 2020, 4:02 AM

#

huh okay

cloud walrusBOT Mar 18, 2020, 4:02 AM

#

Namington:

📎 105878844061286400.png

ripe crest Mar 18, 2020, 4:02 AM

#

surprised considering it seems kinda restrictive

scarlet estuary Mar 18, 2020, 4:02 AM

#

\ses{A,B,C}

ripe crest Mar 18, 2020, 4:02 AM

#

the image of the previous map has to be the kernel of the next map or something

scarlet estuary Mar 18, 2020, 4:03 AM

#

yeah but that restriction should make intuitive sense

ripe crest Mar 18, 2020, 4:04 AM

#

A -> B -> C -> D

#

we would need |B| > A if the map from A to B is injective, otherwise we just get the 0 map

#

right?

oblique river Mar 18, 2020, 4:04 AM

#

if you want the map A -> B to be injective you need to put a 0 before it

#

0 -> A -> B

ripe crest Mar 18, 2020, 4:05 AM

#

makes sense

cloud walrusBOT Mar 18, 2020, 4:06 AM

#

Daminark:

📎 301641201382064128.png

ripe crest Mar 18, 2020, 4:07 AM

#

Okay, let me look at the defn of exact sequences

bleak abyss Mar 18, 2020, 4:07 AM

#

But yeah so it turns out that exactness of a sequence isn't like, oh any sequence of maps you write down will be exact

#

But they are common and you know a lot when you're able to get an exact sequence

ripe crest Mar 18, 2020, 4:07 AM

#

oh okay, it's just the image and kernel thing i said - nothing more

oblique river Mar 18, 2020, 4:07 AM

#

here is a hot take that I heard once

#

about exact sequences

bleak abyss Mar 18, 2020, 4:08 AM

#

🍿

oblique river Mar 18, 2020, 4:09 AM

#

so: kernels are like definable things. something is (very losely) definable if it's in the kernel of some map. Like, you define something as having some property which is like singling it out

ripe crest Mar 18, 2020, 4:09 AM

#

So in 0 -> A -> B, we have no information about B right? We just know that A injects into B?

oblique river Mar 18, 2020, 4:09 AM

#

and images are like constructible things. like, you can construct something from somethign else when you have a map and the thing you're constructing is the image

bleak abyss Mar 18, 2020, 4:09 AM

#

silent: yes

oblique river Mar 18, 2020, 4:09 AM

#

and so like, exactness is like "to what extent are definable things constructible"

#

err, I guess being exact would mean that all definable things are constructible

bleak abyss Mar 18, 2020, 4:10 AM

#

Ah I've been wondering what homology was all about

#

😛

oblique river Mar 18, 2020, 4:10 AM

#

yes :P

ripe crest Mar 18, 2020, 4:11 AM

#

huh okay - i cant fully appreciate that now, but itll marinate in the back of my mind

#

@bleak abyss , I did the first part - it's just using the defintions properly - working on second part

#

did second part

bleak abyss Mar 18, 2020, 4:14 AM

#

Nice

ripe crest Mar 18, 2020, 4:14 AM

#

that helped with familiarizing myself with the definitions!

bleak abyss Mar 18, 2020, 4:15 AM

#

But yeah obv we're in the context of like, either groups or R-modules or something reasonable

#

Okay so now

ripe crest Mar 18, 2020, 4:15 AM

#

right - and in our homework we have stuff about exact sequences splitting

#

havent covered that yet

bleak abyss Mar 18, 2020, 4:15 AM

#

The most common type of exact sequence is called a short exact sequence

#

0->A->B->C->0

#

Let's unwrap that a little bit

ripe crest Mar 18, 2020, 4:15 AM

#

so A -> B injective

#

B -> C surjective

bleak abyss Mar 18, 2020, 4:16 AM

#

Yup, and B->C surjective

#

Now the kernel of the map B->C is the image of the map A->B

ripe crest Mar 18, 2020, 4:16 AM

#

that feels weird

#

because in the first isomorphism theorem, we have surjection -> Injection

#

the decomposition

#

here it's reversed

bleak abyss Mar 18, 2020, 4:16 AM

#

Well, be careful, this is actually gonna make sense

#

So first iso says that f:A->B, then A/ker(f) \cong im(f)

#

Well let's say I have a surjection B->C

#

I immediately have an exact sequence

#

0->ker(f)->B->C->0

#

ker(f)->B is just the inclusion

#

What's the image of that? Itself, ker(f)

#

So really what a short exact sequence of the form 0->A->B->C->0 is saying is that C is the quotient of B by A

ripe crest Mar 18, 2020, 4:19 AM

#

C is isomorphic to B/A

#

no

#

C is isomorphic to B/ker(f) where f:A -> B

#

wait

#

yes

#

C is isomorphic to B /A

#

really though?

bleak abyss Mar 18, 2020, 4:20 AM

#

Well

ripe crest Mar 18, 2020, 4:20 AM

#

We don't have B -> C being injective

#

so

bleak abyss Mar 18, 2020, 4:20 AM

#

The map A->B is an injection

#

So I'm just identifying A with its image in B

ripe crest Mar 18, 2020, 4:20 AM

#

ahhh ... then we mod out by A

#

since B -> C is surjective

bleak abyss Mar 18, 2020, 4:21 AM

#

So yeah A is now a subset of B, and the point is that exactness tells us that A is the kernel of the map B->C

ripe crest Mar 18, 2020, 4:21 AM

#

then we get an isomorphism

bleak abyss Mar 18, 2020, 4:21 AM

#

And the image of that map is C

#

So B/A = C

ripe crest Mar 18, 2020, 4:21 AM

#

huh

#

cool

#

this flattens out our otherwise triangular commutative diagram

#

oh i shouldve looked at the homework .. they have that as an example

#

0 -> K -> M -> Q -> 0

#

and it says K for kernel, M for module, Q for quotient

bleak abyss Mar 18, 2020, 4:23 AM

#

Nerd

ripe crest Mar 18, 2020, 4:23 AM

#

lmao

#

ned

bleak abyss Mar 18, 2020, 4:24 AM

#

Anyway yeah so the idea behind a splitting is that there's something called the splitting lemma

#

0->A->B->C->0, let's say we're in the context of like, R-modules

#

In that R = \mathbb{Z}

#

Well, I have the mantra that B/A = C

#

Now this doesn't mean that B = A\oplus C

#

😢

#

Obvious example being, 0->Z/2->Z/4->Z/2->0 (I'll let you fill in the maps)

#

Well, the splitting lemma says that, for an SES of R-modules 0->A->B->C->0, TFAE:
(1) The map A->B has a retract (a map B->A such that A->B->A is the identity on A)
(2) The map B->C has a section (C->B->C)
(3) B splits naturally as A\oplus C

woven delta Mar 18, 2020, 4:42 AM

#

I like referring to both retracts and sections as sections

ripe crest Mar 18, 2020, 4:43 AM

#

what is \oplus?

woven delta Mar 18, 2020, 4:43 AM

#

The direct sum

ripe crest Mar 18, 2020, 4:43 AM

#

TFAE means?

woven delta Mar 18, 2020, 4:43 AM

#

The following are equivalent

ripe crest Mar 18, 2020, 4:43 AM

#

Ah okay

#

Oh - so if we can split an exact sequence, it means we can factor B very nicely

#

It's sorta analogous to CRT

woven delta Mar 18, 2020, 4:45 AM

#

No

ripe crest Mar 18, 2020, 4:45 AM

#

Actually

woven delta Mar 18, 2020, 4:45 AM

#

I mean I guess

#

I don't like that analogy

#

It's analogous to rank nullity in vector spaces

#

Well it's a generalization of rank nullity

ripe crest Mar 18, 2020, 4:46 AM

#

oh, but we have no notion of dimension

woven delta Mar 18, 2020, 4:46 AM

#

We have a notion of rank actually

ripe crest Mar 18, 2020, 4:47 AM

#

at least i havent learned of one for modules yet

#

If we can write B = A \oplus C, then isn't that similar to writing R[X]/(X-1)(X-2) as R[X]/(X-1) x R[X]/(X-2)?

woven delta Mar 18, 2020, 4:49 AM

#

I mean it's analogous in the sense it looks kinda the same

#

But it doesn't have the same feel to it

#

The ring story vs the module story feels very different

bleak abyss Mar 18, 2020, 4:51 AM

#

Uh, rank nullity?

#

Splitting lemma isn't really much of a thing in vector spaces because lol memes

woven delta Mar 18, 2020, 4:52 AM

#

What do you mean dami?

bleak abyss Mar 18, 2020, 4:53 AM

#

I mean idk in the context of vector spaces

#

Splitting lemma is true

woven delta Mar 18, 2020, 4:53 AM

#

In vector spaces it always holds

bleak abyss Mar 18, 2020, 4:53 AM

#

But turns out I have a much easier criterion for things to split

#

Namely existing

woven delta Mar 18, 2020, 4:53 AM

#

But the way you prove it always holds

#

Is by exhibiting a section

#

I guess my point was that you should think of splitting as being similar to rank nullity

bleak abyss Mar 18, 2020, 4:55 AM

#

I got that, just like, I think it'd be more useful to like, draw an analogy to a context where splitting isn't automatic, you know?

woven delta Mar 18, 2020, 4:55 AM

#

Oh sure

ripe crest Mar 18, 2020, 5:06 AM

#

err okay

#

I know that Magma -> Semigroup -> Monoid -> Group -> Ring -> Field (and we have division rings and integral domains between rings and fields)

#

Where do vector spaces and k-algebras fit in to this? I know a vector space is a field acting on an abelian group. I know a k-algebra is a field acting on a ring (or there's a field in the center of the ring).

#

and I know modules are rings acting on abelian groups

#

so it seems like we have two concurrent streams

#

the one I mentioned above and

#

modules -> vector spaces -> algebras

#

And why do we say R-module, k-vector-space, and k-algebra?

#

why give importance to the thing that's doing the acting as opposed to the thing being acted upon?

#

going to bed soon, so if someone wants to respond, just @ me and I'll look in the morning

scarlet estuary Mar 18, 2020, 5:14 AM

#

i honestly think you're too concerned with "chains" or "heirarchies"

#

like im not sure why rings are the logical next "step" after groups

#

why rings instead of abelian groups?

#

rings introduce an entire new operation, whereas previously we'd only been adding properties to our first operation

#

indeed, one can "think of" rings as sets of endomorphisms over a group, in the sense that there is a ring morphism between (A, +, *) and (Hom(A, +), +, function composition)

#

specifically a -> (x -> ax)

#

from this lens, rings are a totally separate object

#

and i'd argue this is the more natural sense

#

even if it's a bit harder to explain in an algebra 1 lecture

#

anyway, i think the notion you're trying to convey is

#

a "chain" of "more specific forms" of a structure

#

we often represent these as subset chains

#

like, abelian groups $\subset$ groups $\subset$ monoids $\subset$ semigroups $\subset$ magmas

cloud walrusBOT Mar 18, 2020, 5:18 AM

#

Namington:

📎 105878844061286400.png

scarlet estuary Mar 18, 2020, 5:18 AM

#

(where here, each word represents the corresponding set)

#

you might've seen something similar in analysis too

📎 unknown.png

#

anyway, from this framing, you do indeed have algebras subset vector spaces subset modules

#

well

#

actually thats kind of a weird thing to say

#

since the algebra-vector space thing is comparable to the ring-group thing

#

i.e. we're introducing another operation

#

to address your second point: because it's more convenient, in practice, to talk about "the thing doing the acting"

#

like if you've ever taken a linear algebra class

#

you can do proper linear algebra without ever defining a group

#

but you cant do it without defining a ring

#

(or at least a field, but at that point just define a ring too)

#

@ripe crest

#

i shamelessly stole this from a reddit comment, but this is a common "philosophic framing" for these objects

📎 unknown.png

#

not sure whether its helpful to think of them like that right now

#

but these are the conceptions that come up in representation theory and commutative algebra, respectively

red imp Mar 18, 2020, 5:28 AM

#

Hey, could someone give me a hint as to how we can reduce part (d) to only checking 3 cases?

📎 unknown.png

somber bramble Mar 18, 2020, 10:37 AM

#

in class we quickly used the following statement:

If $M_{\alpha}$ are $R$-modules, $N$ is a finitely generated free $R$-module, then $(\prod M_\alpha) \otimes_R N \cong \prod(M_\alpha \otimes_R N)$

is there an easy counterexample to show that freeness is required here?

cloud walrusBOT Mar 18, 2020, 10:37 AM

#

Sascha Baer:

📎 131832027082260480.png

wind steeple Mar 18, 2020, 10:59 AM

#

Maybe Z^N (x) Q provides the subset of Q^N of sequence with bounded denominator. But it is =/= Q^N

somber bramble Mar 18, 2020, 11:17 AM

#

as in $\Z^\N \otimes \Q$ you mean?

cloud walrusBOT Mar 18, 2020, 11:17 AM

#

Sascha Baer:

📎 131832027082260480.png

wind steeple Mar 18, 2020, 11:28 AM

#

Yes

#

Z^N = the infinite product over N of Z

halcyon siren Mar 18, 2020, 12:50 PM

#

📎 IMG_20200318_125007.jpg

upper pivot Mar 18, 2020, 12:51 PM

#

,rotate

cloud walrusBOT Mar 18, 2020, 12:51 PM

#

📎 476552746581884958.png

halcyon siren Mar 18, 2020, 12:51 PM

#

Thank you

#

Err so you see it says adZ restricted to g0, x raised to the power of the dimension is non-zero, why do we need that

#

Isn't the adZ restricted being no singular enough since it guarantees the dimension of dim g0, Z is less than dim g0, x, because the 0 eigenvalue has multiplicity less than dim g_0,x

hot lake Mar 18, 2020, 12:55 PM

#

uncrop

halcyon siren Mar 18, 2020, 12:56 PM

#

That is the entire proof, I'll show you the hypothesis on the prev. page

#

📎 IMG_20200318_125727.jpg

hot lake Mar 18, 2020, 1:04 PM

#

what is a Cartan sub-algebra ?

halcyon siren Mar 18, 2020, 1:09 PM

#

Oh wait nvm I think I gotit

#

what is a Cartan sub-algebra ?
@hot lake A self-normalising nilpotent subalgebra in a lie algebra

hot lake Mar 18, 2020, 1:13 PM

#

okay then I don't need to ask what do those words mean

novel dragon Mar 18, 2020, 10:01 PM

#

Yo can i get some help please on this homework

fading wagon Mar 18, 2020, 10:04 PM

#

what have you tried so far and what's the question

#

@novel dragon

novel dragon Mar 18, 2020, 10:05 PM

#

Ok ima send a pic

#

I dont think this is abstract algebra but i just need hel

#

Help

#

📎 image0.jpg

#

Number 1

#

@fading wagon

fading wagon Mar 18, 2020, 10:10 PM

#

@novel dragon It's probably just #prealg-and-algebra

#

@raven kestrel #calculus

raven kestrel Mar 18, 2020, 10:10 PM

#

ty!

leaden schooner Mar 19, 2020, 12:08 AM

#

How do you go about checking if a polynomial is irreducible over a field?

#

Im doing an excerise in Pinter to show that 1 + 2i is algebraic over Q and I got x^2 -2x + 5

#

Obviously this is a minimal polynomial because if the highest degree was 1 then the constant term would be complex

#

but how do you go about less obvious problems?

uncut girder Mar 19, 2020, 12:10 AM

#

Theres a useful theorem called Eisenstein's criterion

#

https://en.m.wikipedia.org/wiki/Eisenstein's_criterion

Eisenstein%27s_criterion

leaden schooner Mar 19, 2020, 12:11 AM

#

oh thank you!!1

#

Im a little confused

#

for my polynomial, there doesnt exist a prime number satisfying the eisenstein criterion

#

So does that imply this polynomial is reducible?

bleak abyss Mar 19, 2020, 12:28 AM

#

Nope

ripe crest Mar 19, 2020, 1:53 AM

#

@scarlet estuary fair enough - but in Rotman's group theory text, at the back, he has a similar hierarchy as the one im describing - i might have even gotten mine partly from there as well

solemn rain Mar 19, 2020, 1:54 AM

#

just an intuition check

#

so a composition series of a group is just a sequence of groups N_0 < N_1 ... <N_k

#

where N_n is normal to N_(n+1)

#

and N_(n+1)/N is simple

#

right?

#

and the group is called solvable

#

if N_(n+1)/N is simple abelian

#

?

#

are these right definitions

#

?

stone fulcrum Mar 19, 2020, 2:14 AM

#

It's a composition series if the N(n) is a maximal normal subgroup of N(n+1)

#

Which is equivalent to N(n+1)/N(n) is simple yeah

#

And it's solvable if the quotients are also abelian

#

Yeah you've got them

solemn rain Mar 19, 2020, 2:17 AM

#

okay coo

#

l

#

i wanna check if i did this problem right

#

Prove that If G is an abelian simple group then G is isomorphic to Z_p for some prime p

#

proof: suppose G is an abelian simple group

#

since G is simple , G must have no nontrivial normal subgroups.

#

and since G is abelian all ssubgroups would be normal

#

hence |G| is odd hence by feit thompson G is ismorpohic to Z_p for some prime p

#

( if |G| = 2 then G is cyclic , hence isomorphic to Z_p )

#

is that right?

#

and if thats rigth

#

how do i do it if i cant assume G is finite

stone fulcrum Mar 19, 2020, 2:22 AM

#

But |G| isn't necessarily odd? Indeed, what if it's infinite?

solemn rain Mar 19, 2020, 2:22 AM

#

idk

#

thats where im stuck ig XD

stone fulcrum Mar 19, 2020, 2:22 AM

#

Hmm, let me think

solemn rain Mar 19, 2020, 2:26 AM

#

sure

#

thanks for helping

ripe crest Mar 19, 2020, 2:30 AM

#

is the image of every submodule a submodule?

#

I feel like no - we would need the homomorphism to be surjective

#

Because ideals are sub-R-modules

solemn rain Mar 19, 2020, 2:30 AM

#

my bad

stone fulcrum Mar 19, 2020, 2:31 AM

#

In a module homomorphism? Yes I believe it is

ripe crest Mar 19, 2020, 2:31 AM

#

huh

#

Guess I need to look at the def

#

again

stone fulcrum Mar 19, 2020, 2:31 AM

#

I believe in any abelian group, if m divides |G|, then G has a subgroup of order m @solemn rain

#

But check that because ehh

solemn rain Mar 19, 2020, 2:32 AM

#

i htink m has to be prime

#

im not sure either lmao

#

nah ur right

#

its given in an exercise to prove

#

im going to try to prove it ig

solemn rain Mar 19, 2020, 3:13 AM

#

does any11 know where can i get more exercises

#

for undergrad group theory

#

?

ripe crest Mar 19, 2020, 3:25 AM

#

Yes

#

Rotman

ripe crest Mar 19, 2020, 5:25 AM

#

why do we care about algebraic numbers?

solemn rain Mar 19, 2020, 5:26 AM

#

what are those

ripe crest Mar 19, 2020, 5:27 AM

#

From my understanding, they're roots of polynomials in C[X]

#

https://en.wikipedia.org/wiki/Algebraic_number

mild laurel Mar 19, 2020, 5:28 AM

#

roots of polynomials in Q[x]

solemn rain Mar 19, 2020, 5:29 AM

#

ig cuz like

#

maybe these roots make cool stuff

#

cool geometries?

#

lmao just an ignorant suggesting

ripe crest Mar 19, 2020, 5:29 AM

#

oh right - roots of polynomials in C[X] would just give you all of C

mild laurel Mar 19, 2020, 5:29 AM

#

And I mean, lots of questions you have are about rational polynomials right

solemn rain Mar 19, 2020, 5:29 AM

#

ig AG studies roots of polynomials

mild laurel Mar 19, 2020, 5:29 AM

#

Also, the algebraic numbers are the algebraic closure of Q, and algebraic closures are important

ripe crest Mar 19, 2020, 5:30 AM

#

And I mean, lots of questions you have are about rational polynomials right
that's true

#

kinda wondering why they're so interesting that we have algebraic nt

mild laurel Mar 19, 2020, 5:31 AM

#

Lots of number theoretical questions can be answered by going to field extensions of Q

ripe crest Mar 19, 2020, 5:31 AM

#

Ah

#

Cuz Frac(Z) = Q

#

?

#

Well, I haven't learned about field extensions yet

mild laurel Mar 19, 2020, 5:32 AM

#

In some sense

#

Like, one question is what primes can be written in the form x^2 + y^2

#

and so it makes a lot of sense to factor as (x + iy)(x - iy) in the field extension Q[i]

#

and study the question there

ripe crest Mar 19, 2020, 5:33 AM

#

Hmm - do you find that question interesting?

mild laurel Mar 19, 2020, 5:33 AM

#

I think its cool

#

but you can generalize this a ton, ton further

#

what primes can be written in the form x^2 + ny^2 in general has a very nice answer

ripe crest Mar 19, 2020, 5:34 AM

#

I guess the question leads to the development of some cool math - but the question itself doesn't seem interesting to me

#

I guess it's a personal thing

mild laurel Mar 19, 2020, 5:34 AM

#

no i don't disagree

#

it was just an example of what you could do with field extensions and algebraic numbers

#

If you think Fermat's last theorem is interesting, that does kind of the same thing

#

just like how you can factorize x^2 + y^2 over Q[i]

#

you can factor x^p + y^p over Q[\omega] where \omega is a primitive p-th root of unity

#

And this was done by Kummer to try to prove Fermat's Last theorem

ripe crest Mar 19, 2020, 5:39 AM

#

Ah - so it is quite useful

mild laurel Mar 19, 2020, 5:40 AM

#

Turns out it worked, but only for a certain subset of primes called the regular primes

#

(funny how you think fermat's last theorem is interesting but the first question isn't)

ripe crest Mar 19, 2020, 5:42 AM

#

Well, I don't think it's interesting

#

I just meant that it's good to know alg nt was used in solving a really big problem

solemn rain Mar 19, 2020, 5:43 AM

#

why do i never hear of analytic NT

ripe crest Mar 19, 2020, 5:43 AM

#

Because it's analytic

#

and analysis is gross

#

who wants to see loglogloglogloglog

solemn rain Mar 19, 2020, 5:43 AM

#

lmao

#

heard ppl are trying to solve riemann hypothesis

#

just because of something with primes

#

so thats cool ig

opal plover Mar 19, 2020, 5:44 AM

#

Dude

mild laurel Mar 19, 2020, 5:44 AM

#

well, you just don't know enough

#

the prime number theorem is analytic number theory

#

and that's a famous one

#

stuff like green-tao as well

#

or dirichlet's theorem

ripe crest Mar 19, 2020, 5:44 AM

#

I'm mostly joking 😛

#

just being facetious

mild laurel Mar 19, 2020, 5:44 AM

#

I know

#

that was mostly in response to momen

ripe crest Mar 19, 2020, 5:45 AM

#

ah

solemn rain Mar 19, 2020, 5:45 AM

#

do ppl generally like

#

alg nt

#

moore

mild laurel Mar 19, 2020, 5:45 AM

#

idk

#

I don't think it's possible to answer that question

#

and i don't think its a useful one to think about

solemn rain Mar 19, 2020, 5:46 AM

#

yea just wanna know why the sterotype

ripe crest Mar 19, 2020, 5:46 AM

#

https://en.wikipedia.org/wiki/Prime_gap#Lower_bounds

solemn rain Mar 19, 2020, 5:47 AM

#

of analysis is gross

ripe crest Mar 19, 2020, 5:47 AM

#

look at those logs

#

I count 7 in the numerator

#

3 in denominator

solemn rain Mar 19, 2020, 5:48 AM

#

loglogloglog

#

'who wants to see loglogloglogloglog'

#

u just ruined a field

ripe crest Mar 19, 2020, 5:49 AM

#

g_n > \frac{c\log n\log\log n\log\log\log\log n}{(\log\log\log n)^2}

#

imagine typesetting that

#

of analysis is gross
cuz you gotta deal with epsilons and deltas

#

You don't get equalities - only inequalities

solemn rain Mar 19, 2020, 5:51 AM

#

oh

#

manipulating stuff

#

in higher math sucks

ripe crest Mar 19, 2020, 5:51 AM

#

obvs take this with a big spoonful of salt

#

im an undergrad who's only analysis exposure is baby rudin

#

functional analysis looks really cool

#

measure theory looks tedious

#

harmonic analysis idk what it's really about. fourier stuff

solemn rain Mar 19, 2020, 5:55 AM

#

i hear functional analysis is p cool

#

complex analysis is p cool too

#

from what i hear

#

i have no exposure to any analyss

ripe crest Mar 19, 2020, 5:56 AM

#

ah gotcha

#

you should take an analysis course!

#

maybe you'll like it!

solemn rain Mar 19, 2020, 5:56 AM

#

nah man im learning all this just for fun

#

if im not getting fun ofit then there is no reason to do it

#

and i dont think

#

i had any fun

tiny pagoda Mar 19, 2020, 5:56 AM

#

I really like non-complex analysis, just the other day I proved that the constant function was smooth, really fun stuff

solemn rain Mar 19, 2020, 5:56 AM

#

tryign to learn rudin

ripe crest Mar 19, 2020, 5:57 AM

#

hahaha

solemn rain Mar 19, 2020, 5:57 AM

#

yea ig thats liek the most 'good' stuff u can ever prove

#

anything higher than this u get a counterexample

ripe crest Mar 19, 2020, 5:57 AM

#

lmao

#

amen

#

topology is cool though

#

might self-study it over the summer

#

and take an algebraic topology course next yr

solemn rain Mar 19, 2020, 5:57 AM

#

i was texactly thinking of this

stone fulcrum Mar 19, 2020, 5:57 AM

#

Real analysis for the fun of it

solemn rain Mar 19, 2020, 5:58 AM

#

i gotta finish with GT first

#

i was thinking maybe topology or like

#

continuing with ring theory then NT

tiny pagoda Mar 19, 2020, 5:58 AM

#

We can know math stuff - everyone

Godel: COUNTEREXAMPLE REEEEE

stone fulcrum Mar 19, 2020, 5:58 AM

#

Topology is a pretty lit time

solemn rain Mar 19, 2020, 5:58 AM

#

really?

#

like the baby topology?

#

or do u mena like

#

the higher topology

#

?

stone fulcrum Mar 19, 2020, 5:58 AM

#

I'm not crazy skilled at it but I like what I know

tiny pagoda Mar 19, 2020, 5:58 AM

#

Topology sounds really fun to me lol

solemn rain Mar 19, 2020, 5:58 AM

#

what do you know

stone fulcrum Mar 19, 2020, 5:58 AM

#

Mostly baby topology lol

solemn rain Mar 19, 2020, 5:58 AM

#

yea

#

the munkres one?

ripe crest Mar 19, 2020, 5:58 AM

#

I got the Willard book

solemn rain Mar 19, 2020, 5:58 AM

#

or the

#

hatcher one

ripe crest Mar 19, 2020, 5:59 AM

#

ive heard its more concise than munkres

stone fulcrum Mar 19, 2020, 5:59 AM

#

Yeah, and not all of it crazy well

tiny pagoda Mar 19, 2020, 5:59 AM

#

Alg Geo also seems cool

solemn rain Mar 19, 2020, 5:59 AM

#

yea but like munkres seems so gentle and loving

#

like first chapter is just meant to teach u set theory

#

which is like

#

so romantic XD

ripe crest Mar 19, 2020, 5:59 AM

#

only first 6 chapters of munkres is general topology iirc

solemn rain Mar 19, 2020, 5:59 AM

#

like it makes you feel safe lmao

ripe crest Mar 19, 2020, 5:59 AM

#

every undergrad math book does that i feel

solemn rain Mar 19, 2020, 5:59 AM

#

and other books just start chapter 1

#

local class fields or whatever

ripe crest Mar 19, 2020, 5:59 AM

#

aluffi algebra book does that

solemn rain Mar 19, 2020, 5:59 AM

#

yea

#