groups-rings-fields

[agile burrow]

But yeah, I think showing the ext group is non zero might involve a little work and understanding certain Hom groups

[formal ermine]

but P projective iff Ext^n(P, M) = 0 for all n >= 1 and modules M?

[agile burrow]

Well actually, to show the Ext group is non zero it suffices to find a non trivial extension

[agile burrow]

Yes, but what does this have to do with this problem?

Q isn't projective, so we don't know what Ext^1(Q, Z) is

[formal ermine]

ah right I'm having trouble inverting the iff in my head, sorry

[agile burrow]

I guess the contrapositive statement is that P isn't projective if there exists a module M and some n >= 1 such that Ext^n(P, M) is non zero

But that's pretty unhelpful here lol

[formal ermine]

how would you even find a free Z module that is bigger than Q

[agile burrow]

Why do you need to find one?

[formal ermine]

I want to start by finding a projective resolution of Q and just go that way

... -> P_0 -> Q -> 0

P_0 needs to be free and "bigger than" Q

[agile burrow]

Hm. Well you can always take Z^Q but this seems a bit

Unwieldy

So it may be wiser to compute this another way

[formal ermine]

the only other way I could think of doing this would be to find a ses 0 -> Q -> B -> C -> 0 where B is projective and then Ext^1(Q, Z) = Ext^2(C, Z) but I'm not sure if this is optimal

[agile burrow]

I keep approximately one injective resolution in mind at all times

Do you know that you can compute Ext with injective resolutions?

[formal ermine]

nope, this is pretty much all I know about ext

[agile burrow]

Well, Q is an injective Z module so maybe part (2) of the corollary could help you

[formal ermine]

is there any nice equivalent condition for injective modules over pids like the one for projective ones?

[agile burrow]

I think they're the divisible modules?

Might have to double check

[formal ermine]

but Z -> Q -> 0 isn't exact

[agile burrow]

Yeah, divisible is equivalent to injective over a PID

[agile burrow]

Yes, but you can put something on the right in place of 0 to make it exact

[formal ermine]

oh, does the corollary also work for les?

[agile burrow]

I'm confused by what you're asking

Start with a short exact sequence involving Z and Q and apply the corollary

[formal ermine]

right

[agile burrow]

There's a straightforward SES to construct

[formal ermine]

0 -> Z -> Q -> Q -> 0?

[agile burrow]

The RHS shouldn't be Q

[slate tide]

Literally the only injective resolution I know lol

[agile burrow]

The cokernel of Z -> Q isn't Q. But like, you know what it is from first iso theorem or something

[formal ermine]

but then how do we apply the corollary if C isn't Q?

[agile burrow]

Ok maybe I shouldn't have said to apply the corollary

But if you look at the induced LES, something helpful will pop out

[formal yoke]

When we talk about a triangle with a hole viewed as a simplicial complex we just remove the 2-simplicie from the simplicial complex right?
So $K$ is for a triangle and $K'$ for a triangle with a hole
\begin{align*}
K&=\bigl{{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c}\bigr} \
K'&=\bigl{{a},{b},{c},{a,b},{a,c},{b,c}\bigr}
\end{align*}

If so what then changes in the computation of the 1st betti number?
Isn't $\text{dim}; \text{image}(\partial_2)$ still $1$ for $K'$? ($\partial_q$ is the q'th boundary map).

[cloud walrus]

ScapeProf

[wraith cargo]

Dim(im(del_2))=0 for the other case

Because it maps to 0

[lethal dune]

there's no non zero C_2 so \del_2 is just zero

[formal yoke]

Isn't this what we are looking at?
[0\stackrel{\partial_2}{\longrightarrow} \mathbb{F}_2^3 \stackrel{\partial_1}{\longrightarrow} \mathbb{F}_2^3 \stackrel{\partial_0}{\longrightarrow} 0.]

[cloud walrus]

ScapeProf

[formal yoke]

Oh wait no thats only between vector spaces and then \partial_0 just maps to 0 always right

yeah nvm

[tired horizon]

Are these 2 the same set ?

Just with different notations

[dim widget]

They are even the same group.

[tired horizon]

Group of invertible elements

Or muliplicative group

Right?

Of a finite field

[dim widget]

yes

[tired horizon]

Aight thanks

[south patrol]

Hm is calculating cyclotomic polynomials by hand a pain

Or do I just suck at long division to use the like $x^n - 1 = \prod_{m \mid n} \Phi_m(x)$ formula

[cloud walrus]

potato

[delicate orchid]

586240877358350341

[south patrol]

lol

I imagine there's not really a more efficient way than recurrence based on that formula though

[delicate orchid]

simply work with exclusively square free numbers

[south patrol]

Okay let's say for square free ones

Lol

Isn't it still a bit of a pain

[lethal dune]

try recurrence

[delicate orchid]

wow!

[south patrol]

Okay sure

lol

[lethal dune]

lol

ok I didn't see the whole question

[dim widget]

It's not a pleasant process unless n is prime.

[lethal dune]

write a program lol

[dim widget]

n being squarefree doesn't really help in terms of the number or complexity of long divisions you have to do if the prime factors are large.

I don't know of a smarter way to do it than just dividing.

[barren sierra]

It's awful yea

Honestly it's worth learning Python and using SageMath for this stuff

As well as for other stuff

[south patrol]

Yeah sure fair enough

I mostly had exams in mind since I had to copute some for a past paper xd

But yeah then I imagine it's just practice long dividing 🙃

Thank

[rustic crown]

idk anything more than mobius inversion for this situation... but doesn't help with the computations at all

[south patrol]

Yeah sure

Eh this is just a hard projlem ig and there are algorithms on the big scale but by hand not much more than recurrence

Hey ho

[rotund aurora]

I think the idea is to use x^n-1= product{d divides n} Phi_d(x)

no?

[south patrol]

Lol ye

[rotund aurora]

ah sorry didnt read that

looks like a pain tbh

computations are probably enlightening, but I haven't done I admit

[dim widget]

I googled it out of curiousity and most "algorithms" seem to just be making the obvious speedups in the order in which you do the divisions.

[south patrol]

Yeah fair I thought that might be the case sadge

[dim widget]

Or work only in very specific situations

[south patrol]

Fair enough yeah

Well thankies anyway

I am go do more past paper questions

[dim widget]

What is the best proof of the Weierstrass preperation theorem?

Discuss.

Bonus points if you can explain it in a simple and intuitive way.

[sonic coral]

If you were to ask a student to show that a group of some order is not simple, would you also give them the prime factorization or leave it up to them to figure it out on the exam?

[formal ermine]

depends on the number tbh

[sonic coral]

It was 253, I figured out what it was eventually but it just seems unnecessary from a learning standpoint

[delicate orchid]

yeah I'd give them it

11*23?? vile

[sonic coral]

at least I have shown that I only need to check up to root253 for possible divisors

[dim widget]

People who don't remember their divisibility-by-23 test from elementary school are the lowest form of people.

[delicate orchid]

tbh the 11 test would be very easy here

2-5+3 = 0

[sonic coral]

i had no clue that’s how that worked

atleast it was easy to show once I had the prime factorization though

[dim widget]

My strategy with these things is to try 2, 3, 5, 11, then give up.

[sonic coral]

no 7?

[coral spindle]

There is a way to produce fairly easy divisibility tests for numbers coprime to 10

[dim widget]

It's a common misconception that 7 is a prime number, don't feel embarassed.

[sonic coral]

don’t gaslighting me i’ll fall for it

[south patrol]

7 is the smallest prime

[sonic coral]

2 is the oddest prime

[dim widget]

Yeah I just don't remember the divisibility test for 7, it's not difficult to rederive these things but realistically I would just google it.

[south patrol]

Lol tfw they say "compute the Galois group" and the answer is like S3 x GA(1,5)

[coral spindle]

If $a$ is coprime to $10$, we have $pa + 10q = 1$ for some values $p$ and $q$.
Suppose we want to figure out if $10x + y$ is divisibile by $a$.

[cloud walrus]

Boytjie

Boytjie

Boytjie

[coral spindle]

This trick makes it very easy to remember! Since 7*3 = 21, for 7 we have q = -2.

It can be a little frustrating to remember which of p and q you need to use, but it's way better than remembering every q for each value of a.

[rotund aurora]

you have do do bezout first

for this cases, its easier to just write the decimal expansion down, and reduce 10 mod whatever

[coral spindle]

That's easier than it may seem. You only need to work with the last digit.

[rotund aurora]

and you dont need to remember anything

[coral spindle]

e.g. with 13, since I know 7*3 = 21, then 13*7 = 70 + 21, so we have q = -9.

[dim widget]

Hmm but I would think that most people have n*7 memorized for n \leq 12.

[sonic coral]

also why were multiplication tables taught up until 12, like why 12

[coral spindle]

12 used to be a very important number!

[rotund aurora]

because 1+2+3+... =-1/12

[coral spindle]

Well I say used to be

[sonic coral]

LOL

[coral spindle]

If you use the metric system, it's no longer needed (:

[rotund aurora]

also 12 is the number with the most factors

probably

[formal ermine]

my favorite prime is 57

[dim widget]

Indeed I proudly memorized all of my times tables from \zeta(-2) to -1/\zeta(1).

[vagrant zinc]

guys a little help
I have a binary operation defined by
a*b=c
where c is the smallest integer greater than a and b

if I take a=3 and b=7 then c=8 ?

[coral spindle]

24 BTFO

[formal ermine]

mfw 4999 is prime

[coral spindle]

yes

[vagrant zinc]

I can also say that it is commutative

[coral spindle]

Does it depend on the order?

[vagrant zinc]

I am in Z^{+}
and the operation is closed

[rotund aurora]

the sqroot is smoler than 75

u can do by hand

[formal ermine]

it looks so non prime

[rotund aurora]

tbf looks nonprime to me

but im biased probably

because 9 and 4 are composite af

[formal ermine]

I mean

49 isn't prime

499 however is prime again

[south patrol]

49 is prime

[formal ermine]

49999 is prime

[dim widget]

What would you guess about 49999999

[south patrol]

Only divisible by 1 proper nonzero factor

[formal ermine]

I wonder if there's a pattern here

[south patrol]

x^2 + x + 41

[chilly radish]

9 is a prime tho

[south patrol]

If there were a pattern then this would be a better proof of Euclids theorem

Lol

[dim widget]

49999999999 you can see right away if you're smart is divisible by 10183299389 so that one is maybe a bit more obvious.

[south patrol]

Ah okay

Thanks

[dim widget]

If you haven't worked out the divisibility rule for 10183299389 not my problem tbh

[coral spindle]

It's very easy actually! q = 1018329939

[vagrant star]

is that 5 * (10^10) -1

[coral spindle]

Because the number is -1 mod 10 :)

See, the method is sound!

:P

[formal ermine]

hmm.. if our number begins with 4 and ends with n - 1 9s then the numbers which are prime up to some random number I typed in are:
[3, 4, 5, 7, 15, 55, 211, 391, 595]

there is an oeis sequence for this

https://oeis.org/A295988

[agile burrow]

Omg it's like the Fibonacci numbers if you replaced some of them

[coral spindle]

It's like the catalan numbers too, just completely different

we've really hit on something

[formal ermine]

I think if we exchange a countable infinite amount of numbers in the sequence we get that their difference will be equal to the coefficients of the taylor expansion of sqrt(1 - x^2)

pi must be hiding here somewhere

I smell it

[formal ermine]

I'm having issues guessing what the ... here should be

for the proof of the fundamental theorem of symmetric polynomials

m is the maximal monomial in f wrt <_2

is it just e_1^(m_1 - m_2) e_2^(m_2 - m_3) e_3^(m_3 - m_4) etc?

[formal ermine]

yeah that's probably it, just did the algo on a poly and it worked

[dim widget]

I would guess that it's \cdots

Good on you TeXing up your notes

[formal ermine]

I hate you

anyway

I broke wolframalpha

does someone here have mathematica

[formal ermine]

need to verify that this is x^4 + y^4 + z^4

ok fuck apparently I did something wrong

lemme check

oop

I fat fingered a 4 instead of a 2

[dim widget]

Need to work on your micro if you're trying to be a mathematician

[formal ermine]

ong if they're gonna make me calculate symmetric polynomial decompositions in the exam I am actually gonna cry

this is only doable with wolframalpha

[white oxide]

is it the right idea to use lagrange for the second half of the proof?

[static temple]

?

[glacial moss]

what's the easiest way to identify a lie algebra with known simple lie algebras, if you only know the structure constants of your lie algebra?

[south patrol]

I'm not too sure how Lagrange helps

Orbit stabiliser

[white oxide]

orbit stabiliser?

[dim widget]

Yes that is a good idea.

[rotund dragon]

you could also consider taking two odd permutations and compose them

[south patrol]

I was responding to a joke (in a jocular fashion) but the point was that Lagrange's theorem can be seen as a special case of a more general theorem

[static temple]

I dont think its an easy task, first I guess you gotta decompose with the radical and the semi simple

[south patrol]

I wouldn't use Lagrange but perhaps I am just being weird aha

[chilly radish]

You can also try using ||the first isomorphism theorem||

[chilly radish]

I think this is slickest

[south patrol]

Well I was gonna say use uh

|| (# A_n H )( # A_n \cap H) = # A_n # H ||

[chilly radish]

Ah that's also nice

[dim widget]

I think it's all basically the same, it's a pretty dry problem.

[chilly radish]

I like mine better tho

[south patrol]

Actually uh

Yeah nah I mean ||sgn: H -> {+1,-1}|| innit lol

which i assume is what shin meant

[static temple]

You can also use the sylow theorem i think

[south patrol]

Uh now that is overkill lol

[chilly radish]

Yea that's what I meant potato

[white oxide]

so i wrote mr = n! where m is the order of H and r is some positive integer, and i'm supposing for contradiction that the number of even permutations in H is not half of the elements, but i'm a little stuck

would appreciate a hint for my slow ass

[dim widget]

Oh definitely was not imagining that you would use Lagrange in that way

I think it's better to use lagrange on the subgroup of even permutations within H.

If you don't see why that's a subgroup, maybe that's the first thing to prove

[white oxide]

oh yea that was the obstacle i think

i was like wait this only makes sense if the even permutations are a subgroup of H

ok i'll try to prove that

[south patrol]

Hm sps I have a $k$-algebra $A$ (associative, not necessarily commutative) s.t. $_A A = V_1 \oplus \dots \oplus V_n$ for pairwise-non-isomorphic, simple, one dimensional $A$-modules $V_i$. I hope I'm correct in thinking that $A \simeq k^n$ as rings, right? Like as rings $A^{op} \simeq \mathrm{End}A(A) \simeq \prod{i=1}^{n} \mathrm{End}A(V_i) \simeq \prod{i=1}^{n} k$

[cloud walrus]

potato

[south patrol]

Where at the last step I used the fact that V_i is dim 1

[lethal dune]

I mean for any positive integer you can find a prime starting with that

Even 69420

[chilly radish]

I dream of a world where you wouldn't have to specify an Algebra is not commutative

[vagrant zinc]

$ab=2^{ab}$\
$(ab)c=\left( 2^{ab} \right)c=2^{(ab)c}=2^{a(bc)}=a(2^{bc})=a(b*c)$

Is the associativity OK?

[cloud walrus]

Awuita Fria

[chilly radish]

Also yes potato that's correct

[vagrant zinc]

[lethal dune]

No

Check again

[south patrol]

Thanks

Is there anything more you can say about the A-module structure? Or nah

And hm is this the way you would do it too lol

[chilly radish]

Yea pretty much

Unless i'm being stupid

[south patrol]

Only thing is like

Nah nvm lol

[lethal dune]

why such reaction det

[chilly radish]

We did something similar in noncomm, we proved that simple M_n(D) modules for D a division algebra are of the form D^n

And we used a similar proof

Here the ring structure should also be preserved

Actually wait

[formal ermine]

this exercise is really weird and I don't even understand what they are asking me. I am supposed to show that the discriminant of a quadratic is b^2 - 4ac by first reducing delta f = a^2(x_1 - x_2) to its elementary symmetric polynomial form and then use e_1(x_1, x_2) = -ba^(-1), e_2(x_1, x_2) = ca^(-1). I don't get how I'm supposed to reduce it to elementary symmetric polynomial form cuz I get e_1^(-1) on the second step (and so does wolframalpha). I also don't understand how they pulled those e_is out of their ass

[white oxide]

tfw you don't feel like saying trivial

[chilly radish]

The direct sum decomposition needn't preserve the multiplication no? @south patrol

[lethal dune]

Sure D^n?

[chilly radish]

So that the isomorphism is not as rings

[south patrol]

Uhhhhh

[chilly radish]

That wasn't on purpose lmao

[chilly radish]

Like why does End_A(A) decompose as a direct product of rings

As a direct sum of modules sure

[dim widget]

I think the idea is that the modules are nonisomorphic.

[formal ermine]

I don't understand where this fails either, cuz a^2(x_1 - x_2) is obviously a symmetric polynomial so the fundamental theorem tells us this is possible

[south patrol]

So what I meant was yeah like

[chilly radish]

Why would that change the fact that the ring structure needn't be compatible with the direct sum

[south patrol]

An A-equivariant map A -> A just will send each V_i -> V_i

necessarily by multiplication by some scalar

[dim widget]

Because End_A(A) is that direct sum as matrix rings

Maybe write down an example.

[chilly radish]

Hmmm yea ok you're right

My b

So yea this should work

[south patrol]

So like given $f: A \to A$, it is given by like $(v_1,\dots,v_n) \mapsto (\lambda_1 v_1,\dots,\lambda_n v_n)$ yeah and then that gives an isomorphism between $k^n$ and $\mathrm{End}_A(A)$ I suppose

[cloud walrus]

potato

[chilly radish]

Yes of course

I forgot how semisimple stuff works for a second

[dim widget]

So all is well, we are back in the warm embrace of commutative rings.

[south patrol]

Lol

Well the original problem was showing that A was a commutative algebra under certain conditions

[chilly radish]

SAD!

[south patrol]

The exact condition being that it's an n-dimensional k-algebra with n distinct composition factors

[chilly radish]

Artin-Wedderburn gang

[south patrol]

Which immediately reduces to what I gave here

ol

*lol

Yeah I mean it seems these problems are meant to like

Not use Artin-Wedderburn, otherwise I'd have just quoted that lol

[chilly radish]

Yea ofc

It is goated tho

With the sauce even

[south patrol]

It is

[dim widget]

I mean this type of argument is similar to what goes into artin-wedderburn.

[chilly radish]

The best proof of wedderburn artin is with idempotents

[south patrol]

How does that go

I mean the proof we used was a bit odd imo

It used idempotents, but then the final bit just ended up being the argument I gave

[south patrol]

That's what led me to this lol

[chilly radish]

You reduce to wedderburns theorem (this also uses idempotents) by taking minimal left ideals, and you prove wedderburn using the fact that for an idempotent in a minimal left ideal e, eRe is a division alg

And you show that R is End_D(K) where K is the minimal.left ideal

And D=eRe

You use semiprimality to find the idempotent iirc

[prime sundial]

ascending chain of ideals in polynomial ring stabilizes

[south patrol]

I think I saw a paper with this lol

[chilly radish]

Our version of W-A had yhe condition of left artinian semiprimr

[south patrol]

Idk what semiprimality is

[chilly radish]

Means P^2=0 implies P=0 for ideals

[south patrol]

Ah

Nice

[chilly radish]

You use this to show that there's a nonzero idempotent

[south patrol]

Man these reps/modules questions are a pain with knowing what I can/cannot assume lol

[chilly radish]

This result is called brauers lemma

[south patrol]

Oh yeah that's what I'd seen

https://www.thebookshelf.auckland.ac.nz/docs/NZJMaths/nzjmaths022/nzjmaths022-01-010.pdf

thank

[prime sundial]

what can i say about the number of elements in Z_2[x] / (p(x)), where p(x) is irreducible of degree n

[dim widget]

It's always the same

As long as by Z_2 you mean Z/2Z or the field with 2 elements.

[prime sundial]

i do yeah

is it just 2^n

thank you

[south patrol]

Oop

[prime sundial]

i hadn't considered that yet, but makes sense

i feel like a lot of these small facts with polynomial rings are so easy to forget
in group theory stuff made sense and was intuitive, but i'm having to work really hard to remember a lot of these little details about rings and fields

i can understand the proofs for things, but most of this content doesn't feel very intuitive

[formal ermine]

for me it's the other way around

[south patrol]

Okay so my WiFi didn't load properly

Didn't see TTEG had already replied lol

[next obsidian]

TTegga

[rustic crown]

chmet

[formal ermine]

Fuck I'm an idiot

why should that be symmetric

ok yeah this exercise is really really weird

[steel light]

So our new function Sf is the sum of every permutation of f?

Yeah that's what it says okay

And this would be symmetric because if we took a permutation of Sf, then the inputs of the component functions sigma(f) get permuted again, but since we've added together all the possible permutations then sigma(sigma(f)) is just going to be equal to a different sigma(f)

If that makes sense lol, not the best language

[coral shale]

mrowl

[steel light]

Does that mean no?

[south patrol]

Yeah sure feather

Hm that is an interesting use of "k-linear" lol idk

[steel light]

I'm pretty sure I get the idea

Basically it's like

[south patrol]

Well it's fine I think here but it's just k-linear has another common meaning lol but sure

Feather are you comfortable with group actions and stuff

[steel light]

I just learned about them ^_^ so probably not. But the definition seems easy enough

[south patrol]

Ah okay well tbh you will understand what I mean I'm sure anyway but the point is like

Well this is perhaps repeating stuff from the notes, idk the details of what you're reading, but the point is that this map like $f \mapsto \sigma f$ is itself linear, like $\sigma (f + \lambda g) = \sigma f + \lambda \sigma g$ etc firstly

[cloud walrus]

potato

[steel light]

Yep

I noticed that too

[south patrol]

Then you can note that for any $\tau \in S_k$ we have $\tau Sf = \tau \sum_{\sigma \in S_k} \sigma f = \sum_{\sigma \in S_k} \tau \sigma f$

[steel light]

It's a linear combination of permutations of your function

[cloud walrus]

potato

[south patrol]

But then, like, multiplication by tau on the left just cycles round elements of S_k

[steel light]

Yep

[south patrol]

so this sum is just Sf once again

[steel light]

That's a much better way of what I was trying to say the "cycles round" part was what I was having trouble with

[south patrol]

Yeah so in fancier language, multiplication on the left by tau is a bijection S_k -> S_k

But yeah I should point out that this is, in a sense, a common thing you do in representation theory and stuff lol

[steel light]

And with Af it's the same idea except when you apply the permutation to every function it swaps the sign of everything

[south patrol]

Yeah exactly

[steel light]

Okay

I'm having a bit of trouble with understanding the tensor product notation

So we take f, with its regular inputs

Then multiply that by g with all the inputs that weren't in f?

So f(x, y) = xy, g(x, y, z) = xyz, so fg(x, y, z)= (xy) * (z)

Noo that can't be right

It's supposed to be 5-linear

[prime sundial]

i can't tell the difference between splitting fields and normal extensions

[lethal dune]

same

[lethal dune]

[steel light]

I know that's wrong but not why

I don't get the notation lol

[south patrol]

Uhh

it'd be like

(a,b,c,d,e) -> abcde

[steel light]

[south patrol]

Why lol

[steel light]

no comprende

[south patrol]

okay so let h = f tensor g

h(a,b,c,d,e,) = f(a,b)g(c,d,e)

=abcde

[steel light]

Right so I think that's what I said here

[prime sundial]

wrong channel

[steel light]

#❓how-to-get-help

But what if g & f take in common inputs

Unless you want to consider them separate ig? Lol

How does this even make sense? Doesn't this mean g takes in (l - k) inputs? But g is l-linear?? 😭

[chilly ocean]

feather...

[steel light]

Yes?

Didn't you say go to abs-alg last time I asked about permutations and stuff

[chilly ocean]

i did say that, but that has nothing to do with this

how many vectors do you think v_{k+1}, ..., v_{k+l} is

[steel light]

...l...

I'm a dummy

LMFAO

But what do you do with the first v_k inputs of g

[chilly ocean]

i don't see any "first v_k inputs of g" here

[steel light]

😭

That's why I was confused originally

The inputs for g start from v_(k+1)

So g's gotta have at least k other inputs before no?

[chilly ocean]

no

you're plugging the vectors v_{k+1}, ..., v_{k+l} into g

g takes l inputs

there are l vectors

[vagrant zinc]

Show that if $G$ is a finite group with identity e and with an even number of elements, then there exists $a\neq e$, in $G$, such that $a*a=e$

[cloud walrus]

Awuita Fria

[vagrant zinc]

any suggestions ?

[steel light]

How are there l vectors in v_(k+1) ... v(k+l)?

oh I'm fucking stupid

[chilly ocean]

[chilly ocean]

think about pairing each element of G with its inverse

[steel light]

So f(x, y) = xy, g(x, y, z) = xyz

So f is 2-linear and g is 3-linear, so f tensor g is f(x, y) * g(x, y, z)?? That's why I'm confused, since that's z(xy)^2, which is definitely not linear. The only way it makes sense is if we consider the first k inputs of g as separate elements

So f tensor g = f(x, y) * g(x', y', z) = h(x, x', y, y', z) = xx'yy'z

[chilly ocean]

f tensor g is the function of 5 inputs given by (f tensor g)(x,y,z,v,w) = f(x,y)g(z,v,w)

[south patrol]

Nice

[steel light]

But what if x = z and v = y

[steel light]

You'd get a nonlinear function in this case

I think I'm just overthinking it

If x = z and v = y then it wouldn't be a tensor product in the first place, just the product of f and g

What makes it a "tensor product" is that we consider overlapping inputs as independent of one another, so it's still multilinear

[south patrol]

You seem to be confused about labelling variables

Like

Actually see what happens when you plug smth into f tensor g

[steel light]

Yeah I see the difference now lol

[vagrant zinc]

look

[steel light]

Is this accurate? It's what makes the most sense to me

[south patrol]

I'm not really sure what you mean by overlapping inputs

[steel light]

fg doesn't even make sense if the domain of f and g are different, whereas the tensor product still does

[south patrol]

Unless you are a physicist

Jk

But like, x y and z are just letters

[steel light]

Like how (x, y) shows up in f and g

Yeah

I get it now zz

pain

[vagrant zinc]

[south patrol]

This is sort of unreadable sorry

I'm not sure what you've done

[steel light]

I don't think the tensor product is necessarily commutative right?

Or no

[chilly ocean]

come up with an example

[steel light]

I think I'm using the words wrong

Agh I don't know what words I'm supposed to use 😭

[chilly ocean]

stop thinking in the discord chat box and think carefully about what it would mean for the tensor product to be commutative

[steel light]

I can't come up with anything more than f*g = g*f lol

[chilly ocean]

"tensor product is commutative" would mean f ⊗ g = g ⊗ f for each f and g

is this true

[steel light]

In my head I feel like it's not but I'm having trouble proving it symbolically

I'm trying to think of the example functions I used

[next obsidian]

Think of a non-trivial linear function

Two different ones

[steel light]

Sure

[white oxide]

what is the use of the second isomorphism theorem?

[steel light]

mfw I can't even figure out how to write the tensor product properly

Using

f(a, u, b) = au + b [3-linear]
g(x, v, z, w) = xv - zw [4-linear]

[chilly ocean]

j?

[steel light]

oops

f ⊗ g = f(a, u, b) * g(x, v, z, w) = (au + b)(xj - zw)
g ⊗ f = g(x, v, z) ...? what...?

[chilly ocean]

i am doubtful of the multilinearity here

[next obsidian]

Definitely don’t think those are multi linear

[steel light]

😭

[next obsidian]

Just pick a linear map

f(x) = 2x

[steel light]

Yes I see why they're not multilinear now

Agh

Okay

[next obsidian]

g(y) = y

[steel light]

f ⊗ g = 2xy
g ⊗ f = y2x??

Which is only commutative if the product operation on x & y is commutative

[next obsidian]

Wait I’m dumb

[steel light]

Me too babe

[next obsidian]

I mean this is over a field

[steel light]

Right

So it should be commutative

But if it's over a not-field then it depends on if the addition & multiplication operations are commutative right?

[chilly ocean]

i think you should only worry about vector spaces and fields for now

[steel light]

Since we're on a vector space it's necessarily commutative, but if we were considering a general algebra then it need not be?

[chilly ocean]

i don't know what that means. algebras are usually taken to be over fields (vector space with multiplication of vectors)

[steel light]

I thought the underlying structure of an algebra only had to be a ring

agh let mego back and reread

I should maybe take notes

o

I think the second definition is what I'm trying to get at

The operations on the underlying ring don't have to be commutative

Just that when we take a scalar in our field it commutes with the product of elements in the ring (I doubt that's the right wording, but what I mean is the homogeneity condition)

Or rather not commutes but like

We can perform scalar multiplication on either element independently before taking their product, and both of those are equal to each other and equal to as if we'd just taken the product first then scaled it

[slate tide]

Lol, this may be a little much, but I remember the second isomorphism theorem being really useful when constructing the spectral associated to a filtration. See Mac Lane's Homology chapter 11 if you so dare.

[next obsidian]

I would highly recommend against this

[slate tide]

LOL

[white oxide]

the fuck

[next obsidian]

My guess is that if you’re asking about the second iso, you shouldn’t be looking into spectral sequences

[chilly ocean]

i will second this guess

[next obsidian]

It’s the least useful isomorphism theorem frankly, but it just sometimes pops up

And it’s good to know one thing is the same as another

[steel light]

So when is this commutative?

[chilly ocean]

feather you are asking the wrong question

[steel light]

What am I supposed to be asking?

[chilly ocean]

please

[next obsidian]

My example was wrong, I realized you can’t produce a counterexample with two linear maps, but only after the fact

[steel light]

OKay but I still don't see what that does for the question of whether the tensor product is commutative or not

[chilly ocean]

it doesn't do anything because it's not a counterexample to anything

try like, a bilinear map on R^2 and a linear function

[steel light]

Okay

f(x, y) = x.y
g(z) = 2z

f ⊗ g = a(x, y, z) = (x.y) (2z)
g ⊗ f = b(z, x, y) = (2z) (x.y)

I feel like such an idiot I don't see anything 😭 am I just writing the product wronggg

Here . is supposed to mean dot product

[chilly ocean]

please write the inputs

[steel light]

Okay

[wraith cargo]

Wait so f inputs vectors and g inputs just elements of the field?

Ah ok i see

I think it will be clear when you write out the elements of the vectors explicitly

[steel light]

Isn't z supposed to be a vector? 💀

[chilly ocean]

you are considering field-valued functions only

[steel light]

Okay

a((1,0), (0,1), 2)) = (1,0).(0, 1) * 4 = 0
b((1, 0), (0, 1), 2)) = 4 * 0 = 0

[chilly ocean]

if f here is defined on R^2 x R^2 and g is defined on R, say, then f ⊗ g will be defined on R^2 x R^2 x R and g ⊗ f will be defined on R x R^2 x R^2. so they couldn't possibly be equal

[steel light]

I probably shouldn't have picked orthogonal vectors

[wraith cargo]

Also wait are you showing that the tensor product is commutative or non commutative

This Convo is confusing

[chilly ocean]

noncommutative

[wraith cargo]

Ah good

[steel light]

OH

f ⊗ g = a(x, y, z) = (x.y) (2z)
g ⊗ f = b(z, x, y) = (2z) (x.y)

But they output the same result no? 😭

But they're not equal because their domains are different

[chilly ocean]

what happened to "please write the inputs"

[steel light]

I tried that and I still got the same thing, let me write it out again

[chilly ocean]

it doesn't matter anyways. f ⊗ g couldn't possibly be equal to g ⊗ f because their domains are not the same

don't bother

[steel light]

Right

Does this mean that f ⊗ g might be defined whereas g ⊗ f might not be?

[chilly ocean]

both are defined, just on different domains

[white oxide]

what do they mean by factoring the map by K?

[chilly ocean]

they mean what is written immediately after

[steel light]

Okay I understand

[white oxide]

... what does that mean

[steel light]

Just wish I had a counterexample where the two tensor products had different codomains

[white oxide]

oh my fault

i read it as YH = YH /K YK

[chilly ocean]

what do you mean different codomains. you're only considering functions going to R

[steel light]

I'm going to get cancer trying to understand this

Sigh

Is the only reason the tensor product isn't commutative just because the domains are different

Or is that just because we're considering multilinear functions over vector spaces

If I look at it with this definition I think I could see why else it isn't commutative

If we swapped a^i and a^j we'd be picking out different elements of v & w

[chilly ocean]

consider
f: R^2 x R^2 -> R given by f(x, y) = x dot y
g: R^2 -> R given by g(z) = z_1 (the first component of the vector z)
then f ⊗ g and g ⊗ f are both defined on R^2 x R^2 x R^2 and, if (u, v, w) is in this, then
(f ⊗ g)(u, v, w) = f(u, v)g(w) = (u dot v)w_1
(g ⊗ f)(u, v, w) = g(u)f(v, w) = u_1(v dot w)
which look not equal

[south patrol]

Why should it be commutative

[steel light]

I don't think it should

I don't think it shouldn't

[south patrol]

You are chucking together two potentially very different functions basically

[steel light]

I SEEEEEE

So would it just not be commutative if our functions weren't symmetric?

[next obsidian]

g is not multilinear

[steel light]

im

gonna pause and drive home and stop thinking about math for a bit

my head hurts

[chilly ocean]

what the fuck are you talking about chmonkey

[next obsidian]

Im pretty sure

Uh

[steel light]

Oh

no it is multilinear isn't it?

[chilly ocean]

it's a linear map on R^2

[steel light]

Yeah

[next obsidian]

I mean don’t you need to have g(a,b+b’) = g(a,b) + g(a,b’)

[chilly ocean]

ok let me write V = R^2. then im considering f: V x V -> R and g: V -> R

[steel light]

If you scale the vector you get the scaled first coord and if you add two vectors the first coord of the sum is the sum of the first coords of each

[white oxide]

why does (Z/6Z)/(2Z/6Z) have 2 elements and not three if half of the elements of Z/6Z lie in 2Z/6Z

[next obsidian]

Ohh

Kekw

I see

[chilly ocean]

lmao

[next obsidian]

I thought you were taking V = R

Lmfao

[chilly ocean]

honestly fair

i could have been a bit clearer

[steel light]

Okay I think I just need to see tensor products in action more to grow more comfortable with them

[slate tide]

Half of the elements of Z/6Z lie in 2Z/6Z so the latter contains 3 elements. Thus the quotient has 6/3=2 elements

[wraith cargo]

When you take the quotient you're gonna get two cosets
One coset contains 6Z, 2+6Z and 4+6Z, while the other one contains the other three odd integers

[white oxide]

ye that makes sense, but aren't (1 + 6Z) + (2Z + 6Z), (3 + 6Z) + (2Z + 6Z), (5 + 6Z) + (2Z + 6Z) three elements tho

sorry im lost

need a map

[wraith cargo]

i.e. your cosets are gonna look like 2Z/6Z and 1+(2Z/6Z)

[wraith cargo]

The last element is the same as the first

[white oxide]

oh right

ohhhhh

ok thanks got it!

[steel light]

I see now

So the tensor product on its own would just be

f(v1, v2)g(v3)

But once we apply the operator A, we take the linear combination of permutations of the inputs

This looks strangely like the Leibniz identity

I understand derivations act on functions whereas here we're just switching inputs

They just look kinda similar lol

[prime sundial]

any hints?

[wraith cargo]

Contradiction

||i.e. take some polynomial in α and assume it's algebraic and reach a contradiction||

[next obsidian]

That’s not enough, not every element is of that form

[wraith cargo]

Ahh true

Brain tired

[vagrant zinc]

name of book?

[white oxide]

fraleigh

[vagrant zinc]

do i have it in physical form page number?

[white oxide]

my fault boss, lemme get back to you on that stopped studying

[prime sundial]

since the degree of alpha is n, then we have a minimal polynomial of degree n, and so at most n roots which we can permute with conjugation isomorphisms
is that the gist of it

[south patrol]

Yes, good

[prime sundial]

nice!

[south patrol]

:)

[prime sundial]

finally did one without needing like 5 hints

[south patrol]

This is really key for Galois theory

Idk if you're doing Galois theory rn though

[prime sundial]

we got to it

[south patrol]

But I can give you a taste lol

Ah okay

[prime sundial]

my final is tomorrow and the prof said we probably will have some galois stuff in there but he didn't give us hw or anything on it so idk what to expect

[south patrol]

Ah okay

[prime sundial]

maybe the correspondence between the klein 4 group and Q(sqrt2, sqrt3)

it's like the main example we've hammered in over and over again

[sonic coral]

I hate when they throw content you havnt been tested on before, onto the final

[south patrol]

Indeed that proof in fact works for F(sqrt(a), sqrt(b)) whenever a,b,ab are all squarefree in F, F not characteristic 2

[prime sundial]

oh nice

i'll keep it in mind

[south patrol]

Bt yeah idk

[terse rose]

The question was to find how many elements of each order there are in G

They considered each element of G and applied the lemma of |<g^k>|=12/gcd(k,12)

My question is take for example the elements of order 12,is the cyclic subgroup of each of them the same as the Group G

And also what other ways could you solve this by? Using la grange’s theorem?

[dim widget]

Yes because the group has only 12 elements.

[dim widget]

All ways boil down to noticing that the group is cyclic of order 12. There is a formula for the number of elements of order m \leq n in a cyclic group of order n which characterizes cyclic groups amongst abelian groups.

[frigid lark]

Does Euler Totient function work (probably not easier)

Oh wait, you want specific elements

[toxic zephyr]

are there any simple examples of algebraic objects that aren't associative? something that maybe a high schooler could understand?

[slate tide]

R^3 with the cross product? Right hand rule is very concrete to me

Possibly see this rock-paper-scissors game: https://en.wikipedia.org/wiki/Commutative_magma#A_commutative_non-associative_magma_derived_from_the_rock,_paper,_scissors_game

[jaunty forum]

Describe the Galois group of the field extension $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$. That is, give its order, write down the elements as automorphisms in Aut(L) and determine to which abstract group they are isomorphic

[cloud walrus]

LeftySam

[jaunty forum]

So I'm (pretty sure) the minimal polynomial is t^2 - 2

[formal ermine]

correct

[formal ermine]

is this Aut(L) or Aut_Q(L)

assuming it is the latter, going the other way is "probably" easier

i.e. determining its galois group then finding the automorphisms

[jaunty forum]

Sorry I tried to specify it, because there's another example so to make it easier

[formal ermine]

what is the order of the galois group?

[slate tide]

technically these are the same thing in this case. Every automorphism of a field fixes the prime subfield.

[formal ermine]

right

til the term prime subfield

[jaunty forum]

The order is equal to the number of roots of t^2 - 2 that lie in Q(root(2)), I think? Which is both of them, so the order is 2?

[slate tide]

Yeah the prime subfield is just intersection of all subfields of L, so in this case Q. But in characteristic p, it'll be Z/pZ.

[formal ermine]

I find this definition confusing tbh

I find "the subfield generated by 1" more clear

[slate tide]

That's definitely better lol

[formal ermine]

the order of the galois group is equal to [Q(sqrt(2) : Q] (because the extension is galois), you've found the minimal polynomial of that already... which is t^2 - 2, so the extension has degree 2 and thus also the group

how many groups of order 2 do you know?

[jaunty forum]

Um, Z/2Z I guess

[formal ermine]

exactly

one of those is the identity

find the other automorphism

(hint: automorphisms permute roots, proof: exercise)

[jaunty forum]

I'm confused because the question says the following: Hint: Use the bijection from Theorem 13.3 (iii) (see the proof), I think the bijection it's referring to is Gal(L/K) -> {a in L such that \mu_{b}(a) = 0}. It says do not define maps L -> L and show that they are field automorphisms, this will be way too much work

How does that bijection come into play here?

[dim widget]

What you just said (what the hint is eluding to) is exactly what illum said

[chilly ocean]

are there any simple examples of algebraic objects that aren't associative?
lie algebras my beloved

something that maybe a high schooler could understand?

[formal ermine]

are you asking for simple and intuitive examples?

[slate tide]

Again, the cross product on R^3 is both a lie algebra and awesome

[dim widget]

What if the high schooler comes up to you and says “but isn’t the Jacobi identity just another incarnation of associativity?” Wyd then?

[jaunty forum]

what does automorphisms permute roots mean in this context

Also the "proof: exercise" is triggering to me because so many of my lecturers just leave so much as exercises lmao

[formal ermine]

if alpha is a root of f then so is sigma(alpha) for any sigma in the galois group

[formal ermine]

yeah but this one is a really easy proof

it's just definition chasing

[jaunty forum]

So, for instance, here; does phi(root(2)) = root(2) and let's say sigma(root(2)) = - root(2) work?

[slate tide]

what if I told you the only thing I know about lie algebras is that the cross product on R^3 is a lie algebra and pauli matrices generate su(2)

[formal ermine]

phi is the identity

sigma works

[elder wave]

calling them a nerd

immediately

[formal ermine]

https://tenor.com/view/cat-addiction-cocaine-snow-funny-gif-24451996

[jaunty forum]

Okay, nice.

Is it cool if I ask another?

[formal ermine]

sure

[jaunty forum]

Basically do everything we just did but for F_2(alpha)/F_2 where alpha has minimal polynomial t^3 + t + 1

[formal ermine]

show work

[jaunty forum]

Wdym?

[formal ermine]

show what you've done/tried already

[jaunty forum]

I haven't really made any progress with this one tbh

Apart from noting the Galois group's order is at most 3

The following Theorem may be relevant: If K is a finite field with p^n elements then the map
Φ: K → K; x → x^p
is an element of order n in the Galois group Gal(K/F_p). Furthermore, the group Gal(K/F_p)
is cyclic with order n and a generator is given by Φ, i.e., Gal(L/K) = ⟨Φ⟩.

[white oxide]

can somebody give me a hint for this question? im assuming that the equality g^5 = e does not hold for all g in G and am trying to use that to prove that G is cyclic

nvm i got it

[south patrol]

Poggers

[white oxide]

is it normal to spend days on a section

or am i just stupid

like i have to read it over and over again and take notes

and then move on to the problems

cuz i really want to solidify the material

[slate tide]

I've spent days on like 2 pages of something lol

[lethal dune]

You spend days on a single line sometimes

[slate tide]

So true

[white oxide]

damn ok that's good to know

[formal ermine]

wtf antihomomorphisms

[chilly ocean]

indeed

[formal ermine]

that is such a weird name

I would just call them like

[lethal dune]

smsihpromomoh?

[formal ermine]

opposite homomorphism or something lol

[chilly ocean]

that's so long

antihomomorphism is fine

[lethal dune]

or the one I suggested

same length

[formal ermine]

just call it opphomomorphism

1 less letter than antihomomorphism

😎

[chilly ocean]

because it's clear what "opp" means in "opphomomorphism"

[rotund aurora]

anisotropic homomorphism

[formal ermine]

I had to look up what antihomomorphism means

[dim widget]

Cohomomorphism

[lethal dune]

No

[rotund aurora]

btw, antihomomorphism is f(xy)=f(y)f(x) ?

[chilly ocean]

way back when i was taking a class on lie algebroids the prof started talking about "vector bundle comorphisms" and it was some cursed shit like "map E_2 -> E_1 covering a map M_1 -> M_2"

[formal ermine]

yes

antihomomorphism from X to Y is a homomorphism X -> Y^op

[rotund aurora]

i think I have seen the prefix "anti" used too when theres like some ordering or a lattice (and the ordering is reversed)

like in a Galois connection

[dim widget]

For groups it's a dumb notion but in general it's interesting.

[formal ermine]

what about rings

ALLRINGSARECOMMUTATIVEALLRINGSARECOMMUTATIVEALLRINGSARECOMMUTATIVEALLRINGSARECOMMUTATIVEALLRINGSARECOMMUTATIVEALLRINGSARECOMMUTATIVEALLRINGSARECOMMUTATIVEALLRINGSARECOMMUTATIVEALLRINGSARECOMMUTATIVEALLRINGSARECOMMUTATIVEALLRINGSARECOMMUTATIVEALLRINGSARECOMMUTATIVE

[rotund aurora]

xdd

[formal ermine]

show me a ring that isn't commutative.

I dare you

[rotund aurora]

your mum

[formal ermine]

touché

[chilly ocean]

matrix rings

stop shitposting

[dim widget]

M_n(D) where D is a division algebra.

[formal ermine]

prove that it exists 🤓

[agile burrow]

Z[S_3]

[formal ermine]

anyhow

[dim widget]

But it's a well known theorem that ryu taught me that besides products of those there isn't anything else.

[lethal dune]

[chilly radish]

comomorphism

[lethal dune]

sry who taught?

You forgot to include semisimple

[chilly radish]

you also need left artinian

for wedderburn artin

or right artinian equiv

[dim widget]

rings are always artinian

[lethal dune]

ye

[wraith cargo]

Sees theorem with artin in it's name
Leaves

[dim widget]

Sorry I made a mistake but I edited it

[chilly radish]

I mean it depends on your definition of semisimple

cuz sometimes it means the jacobson radical is 0

but there's defns that imply artinianity

[lethal dune]

nilpotent also fine right? Out of touch with non comm stuff

[chilly radish]

not sure what you mean

you can also require just semiprime tho

meaning P^2=0 implies P=0 for two-sided ideals

[dim widget]

Yeah I would just say "all finitely generated left modules are sums of simple modules" for semisimple

[chilly radish]

so that would imply left artinian

[lethal dune]

this is wrong, nvm just looked up

[chilly radish]

actually being semisimple is equivalent to being side-artinian and having 0 jacobson radical

which is cool

[lethal dune]

yes

[chilly radish]

I think the weakest condition I know for W-A is left artinian and semiprime

[lethal dune]

what's semi prime again?

[chilly radish]

.

[lethal dune]

I see

thought semiprimary

so many similar names

[chilly radish]

Yea lmao

[south patrol]

Semiprime

[white oxide]

why is the multiplicity of b greater than or equal to the multiplicity of a if phi is the identity on F and hence F[x]?

also i'm a bit confused since they wrote (x - b)^m, wouldn't that imply that b also has multiplicity m? why is the paragraph after that expression needed?

[wraith cargo]

So you can write phi(f(x)) = f(x)

this follows since f(x) is an element of F[x]

so all the coefficients are fixed

now we know that phi sends a to b

so we can see that we can decompose f(x) as (x-b)^m times some polynomial in E[x]

now the idea that is stated here is that this shows that b has multiplicity at least m

(which is the multiplicity of a)

but it's not exact since there could be extra factors of (x-b) in that polynomial to the right

[white oxide]

oh yea i guess that's what i'm confused about, doesn't phi also fix the coefficients of g(x) since g(x) is in F[x]?

[wraith cargo]

g(x) isn't in F[x]

it's in E[x]

[white oxide]

wait but isn't that also true for f(x) then

[wraith cargo]

no

f(x) is in F[x]

[white oxide]

ye but E is also an extension of F right, so it's technically in E[x]

ig i'm confused because they said then in E[x] we may write

[wraith cargo]

here it's better to signify that it's in F[x] since that means all the coefficients are fixed by the F-linear isomorphism

what I meant was

g(x) is more generally in E[x], i.e. it's coefficients need not lie in F[x]

[white oxide]

ohhhhh

[wraith cargo]

because if it were in F[x] then that would imply that f(x) has a factor in F[x]

so it isn't irreducible

which is a contracition

[white oxide]

ah that makes so much sense

thank you!!!

[wraith cargo]

np

[white oxide]

is it true that if F is any field and a1 and a2 are some elements in an extension E of F, then F(a1, a2) = (F(a1))a2?

[wraith cargo]

If you mean (F[a1])[a2] then yes

[white oxide]

can i get a hint for this question plss

[wraith cargo]

show inclusions

i.e. LHS is included in RHS and vice versa

[white oxide]

ig the main trouble that i'm having rn is trying to find an irreducible polynomial that has sqrt(2) + sqrt(3) as a root

yea i'm doing that

i tried it in the obvious way

just expanding

[wraith cargo]

OK so I assume you showed that RHS \subseteq LHS

that's the trivial direction

[white oxide]

yea

oh wait my fault

i meant no not yet

working on LHS

subset RHS

[wraith cargo]

Ah ok so you're doing the hard direction first

Actually

I don't think you need to do that

You trivially have that RHS \subseteq LHS

so you have a tower Q \subseteq Q(sqrt(2) + sqrt(3)) \subseteq Q(sqrt(2), sqrt(3))

now I leave the rest up to you

[white oxide]

wait sorry does it just follow easily because any field containing sqrt(2), sqrt(3) and the rationals has to contain sqrt(2) + sqrt(3) and the rationals

for the trivial inclusion

RHS subset LHS

[wraith cargo]

Essentially yes

[white oxide]

hmmm ok thanks

[wraith cargo]

the idea is that sqrt(2) + sqrt(3) is an element of Q(sqrt(2), sqrt(3))

[white oxide]

ah right

[wraith cargo]

so clearly every element of Q(sqrt(2) + sqrt(3)) is also an element of the larger field

[white oxide]

ah ok that makes sense

thx!

[wraith cargo]

yeah now it's a very easy degree argument

np

[white oxide]

just to make sure i'm underestanding correctly, a subnormal series is a finite sequence of subgroups of G that need not all be normal in G, but a normal series is a finite sequence of subgroups of G that are all normal subgroups of G?

ok now i sound kinda stupid for asking that

[white oxide]

based?

[chilly ocean]

chatgpt = automatic cringe

it didn't even get the statement right

the second paragraph is useless

the third paragraph barely functions as an example

[frigid lark]

Where is butterfly + Jordan Holder?

[white oxide]

how important are those theorems btw

cuz I’m going over them rn and they’re wack

well only butterfly

haven’t gone over Jordan holder yet

[frigid lark]

As far as I am concerned, butterfly is only useful for Jordan Holder, which is used for Galois theory, specifically solvability of field extensions

[white oxide]

Oh shit okay

[warm ember]

help

i got that $\mathbb{C}[x]/(x^2-x)={ax+b\mid a,b\in\mathbb{C}}$

[cloud walrus]

Kevin Yang

[warm ember]

but $(a_1x+b_1)(a_2x+b_2)=(a_1a_2+a_1b_2+a_2b_1)x+b_1b_2$

[cloud walrus]

Kevin Yang

[frigid lark]

As rings, or as vector spaces?

[warm ember]

rings

so with the map $ax+b\mapsto (a,b)$ we would expect $(a_1x+b_1)(a_2x+b_2)=a_1a_2x+b_1b_2$ but this is not the case

[cloud walrus]

Kevin Yang

[warm ember]

also with this map, 1 does not map to 1

i cant think of any other map possible

[frigid lark]

Maybe 1 to (1,1) and X to (0,1)

I haven worked through that map though

[warm ember]

wait actually the map $a(1-x)+bx\mapsto (a,b)$ works

[cloud walrus]

Kevin Yang

[warm ember]

i think this is the same thing

thanks

[formal ermine]

crt

[warm ember]

i dont understand this problem

an inclusion-preserving bijection is just an isomorphism?

also arent there multiple ideals of R/I

[chilly ocean]

an isomorphism of what structure?

[chilly ocean]

is that a problem?

the goal is to find a bijection F from the set of ideals of R/I to the set of ideals of R which contain I, such that, if J and J' are ideals in R/I, F(J) is contained in F(J') whenever J is contained in J'

[warm ember]

rings

[warm ember]

oh

[warm ember]

o

[chilly ocean]

what is your proposed ring structure on the set of ideals of R/I?

[warm ember]

thank you

[chilly ocean]

it may be easier to go the other way around, actually

[warm ember]

ill try

[warm ember]

solved

[wooden ember]

Aight I just want to make sure I’m not oversimplifying things here. For b) we’ve shown in a previous exercise that $|sigma_\infty$ is injective and has dense image in $K_\infty$, so since $\sigma_\infty \otimes_\mathbb{Q} \mathbb{R}$ is also injective (cause we’re tensoring over $\mathbb{Q}$) and going to $K_\infty$ as an $\mathbb{R}$ algebra, we get that $\sigma_\infty \otimes_\mathbb{Q} \mathbb{R}$ has image a dense $\mathbb{R}$ sub module, so it must be the whole algebra.

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

(Only interested in b) rn)

Oh and by K_inf I mean R^r1 oplus C^r2

[kind jacinth]

homomorphism G1->G2.
order of kernel G1 is equal to order of a subgroup G2 right?

[dim widget]

No only the ratio $[G_1:Ker]$

[cloud walrus]

Topos_Theory_E-Girl

[kind jacinth]

if p(x) = (x^2-2)^2 which is an element of Q[x]. is Q[x]/<p(x)> a field?

the answer says no but why not? p(x) is irreducible

[dim widget]

It's not irreducible

You just wrote it as the square of a polynomial

That's like me saying why is 3^2 not prime?

[kind jacinth]

so p(x) = (x^2-2)(x^2-2) is this one reducible?

[south patrol]

It was just said this wasn't irreducible

[dim widget]

Please don't use chatgpt to learn math, or at least don't post about it here.

It's genuinely so depressing.

[south patrol]

Yeah

It makes me feel old being more against tech now lol

Back in the good old days when all we had was textbooks written on scrolls

[dim widget]

Chatgpt doesn't have a mind, it doesn't have a soul, it can't help you understand things that you don't understand because it will just spew random syntactically correct-looking nonsense at you that approximates the answer you're asking for. If you have no way of debugging its output it's worse than useless.

Like if I were going to make a tool to sabotage an undergraduate student's understanding of math it would be like chatgpt

Because, like that one annoying first year graduate student, it confidently spews correct-sounding nonsense interspersed with some true facts and syllogisms.

It teaches you to be the exact thing that would be the worst outcome from an education in mathematics: someone who has absorbed the rhythm and speech patterns of mathematical thought but none of the ideas.

[dim widget]

We need a butlerian jihad fr fr

[south patrol]

Mood

[elder wave]

Chatgpt posting is not only discouraged but forbidden

[lethal dune]

Is that classified as shitposting or illegal

[summer path]

Illegal shitposting?

[kind jacinth]

if it isnt irreducible it reduces to what then?

😭

because (x-sqrt(2))(x+sqrt(2)) isnt correct since sqrt(2) isnt in Q

i am unsure what you mean when u react to my comments. Like is what I am trying to say unclear?

[delicate orchid]

(x^2-2)^2 isn’t irreducible because you’ve literally written it as the product of two quadratics. It’s like asking if 2*2 is prime

[kind jacinth]

ahh i see... ok what if we had it just as (x^2-2)?

then its irreducible right? but (x^2-4) is reducible

[delicate orchid]

yeah x^2-2 is irreducible over Q

[kind jacinth]

thanks!

[trail stump]

Is it correct to say that the Jordan-holder for abelian groups essentially boils down to the fundamental theorem of arithmetic?

[dim widget]

No.

Maybe for cyclic groups it comes down to that and the chinese remainder theorem.

[trail stump]

Because since G is abelian, the factor groups of the series are also abelian, and then since abelian groups can be factored to groups of any order kinda like a number factored to primes.

Ok let’s say G is finite

One thing I guess I haven’t thought about is the fact that abelian group even with the same order can be not isomorphic

[hollow tartan]

What's the idea behind gauss proof that every cyclotomic polynomial is irreducible over Q[x]

[dim widget]

What is Gauss's proof?

[tribal moss]

Is there a more direct connection between the naming of "integral domain" and "algebraic integer", other than "each of them generalizes some properties of Z"?

[dim widget]

No

[slim kayak]

It sorta works for finitely generated abelian groups where you can decompose a group into the direct sum of cyclic groups and a free group. Jordan-Holder is more general than that and you will quickly run into the "extension problem", seeing that you can't uniquely recover your group from your composition series, nor do I think there is any way of listing all possible groups with a given composition series.

[slim kayak]

If your polynomial has degree 3 or lower you can check if it is reducible over a field K if it has a root in K. The roots of x^2-2 are pretty famously not rational numbers.

[dim widget]

But why can you decompose as a product of cyclic groups? That's a decent sized part of the classification of finitely generated abelian groups.

And it definitely doesnt reduce to n=p_1^n_1 ... p_m^{n_m}

[slim kayak]

I meant to say that the prime factorization analogy kinda works with finitely generated abelian groups thanks to the structure theorem.

[dim widget]

It's a good analogy but the question was "does the Jordan holder decomposition boil down to the fundamental theorem of arithmetic?"

[slim kayak]

If you interpret "boil down to the fundamental theorem of arithmetic" to mean to mean that your object can be uniquely decomposed and "re-composed" from that unique decomposition, then no. That's what the second half of my msg is about 💀

[dim widget]

As far as I understand the analogy is just that R embeds into its field of fractions, so it's "like an integer ring" in that sense.

[trail stump]

honestly the way I thought about this was something much more naive and simple

😂

@slim kayak but I think when we reconstruct the original group, we just do joint of the groups, which is the smallest group that contains both

[formal ermine]

I may have misunderstood my prof but

why do we need the stffgmoapid

to show that if

( \bZ^n \iso A \subseteq R \subseteq B \iso \bZ^n )

then $R = \bZ^n$

[formal ermine]

(as Z mods)

[dim widget]

Err I mean you don't

But showing that is part of the structure theorem.

[formal ermine]

wdym

[dim widget]

Like inevitably as part of the structure theorem you need to show that rank(A) is well defined for an abelian group A, and that rank(A/H) \leq rank(A) for H a subgroup.

Or not inevitably, but that's a natural part of the proof.

You might also want to show that rank(H) \leq rank(A).

[formal ermine]

what is the rank of a module

[dim widget]

The rank of an abelian group can be defined in many ways.

But the most natural is $\text{dim}{\mathbb{Q}}(\mathbb{Q} \otimes{\mathbb{Z}} A)$

[cloud walrus]

Topos_Theory_E-Girl

[formal ermine]

ic

[dim widget]

The default \mathbb{Q} is so ugly

[delicate orchid]

the rank is the number of 0s in they smith normal form

[formal ermine]

smith normal form?

[delicate orchid]

yup

the statement is true but completely useless as the proof that a smith normal form even exists requires the fundemental theorem of modules over PID

rather circular innit

[hollow tartan]

The one where he proves for some p prime and its Phi_p(x)

[dim widget]

Okay but how does he prove it? I don't know the mathematical history here.

[hollow tartan]

I'm sorry for my ignorance but is this rank similar for a group?

[dim widget]

No there isn't a good notion of rank for a nonabelian group.

[trail stump]

So the direct product of the factor groups of normal series {Hi} of G, is G right?

[delicate orchid]

it's either 1, 2, or uncountably infinite

[dim widget]

No

E.g. Z/p^2 is Z/p \to Z/p^2 \to Z/p

So the jordan holder factors are just Z/p

But (Z/p)^2 has no elements of order p^2

[south patrol]

hm do you mean like Z/p^2 > Z/p > 1

[trail stump]

Wait I think I meant inner semi direct product