waitttt
#groups-rings-fields
fair shard
it should be right
cause the xcac^-1x^-1
and the cac^-1 in the middle collapses to a
but if you have cxax^-1c^-1 idk what you could do with that
unless the centralizer of a is the same as the centralizer if all conjugates of a
I feel like group theory when you're first starting is like the time when the piñata bursts and it's time to pick up all the candy. there are sooooo many patterns and sooooo many symmetries it's like overwhelming and I have a hard time containing my thoughts
thorn delta
if you defined conjugation as h^-1ah then right cosets would make more sense
fair shard
yup
this excites me so much
but also
I'm overwhelmed
I'm thinking about group actions and conjugation and I'm imagining these elements being moved all over the place with permutations and stuff and it's a lot for me at the moment
hey @thorn delta do you have any suggestions for books or places to go to learn algebra and get intuition for it?
fair shard
I mean math can't really be that dry
maiden ocean
check the pinned messages in #book-recommendations
theres like
a huge list of intro algebra textbooks
chilly ocean
d&f is like
maiden ocean
D&F is just kind of 💤
chilly ocean
df is not exciting
maiden ocean
yea
everything is individually fine but the flow is like
not great
and the exposition is also not great
like i feel like if it was the same way but for like AT it would be way less popular
fair shard
i took a look at the list
it kind of kills me that jacobson called it "basic algebra"
why do you need to say that!
maiden ocean
the first 4 chapters of volume 1 are enough for a year long undergraduate course in algebra
thorn delta
tell ur friends "yea, im reading basic algebra atm"
maiden ocean
debatably just ch 1, 2, and 4
fair shard
then whats sl basic about it
maiden ocean
idk why he titled it that way tbh
fair shard
is it basic?
thorn delta
its normal
fair shard
im reading a book rn that i would honestly say is pretty introductory
maiden ocean
volume 2 contains a lot of shit as well
Volume II comprises all of the subjects usually covered in a first-year graduate course in algebra. Topics include categories, universal algebra, modules, basic structure theory of rings, classical representation theory of finite groups, elements of homological algebra with applications, commutative ideal theory, and formally real fields. In addition to the immediate introduction and constant use of categories and functors, it revisits many topics from Volume I with greater depth and sophistication. Exercises appear throughout the text, along with insightful, carefully explained proofs.
fair shard
and after that i think i will do jacobsom
fair shard
the hard thing is everything ive learned so far in math has either been immediately obvious or obvious after a few minutes of thinking
algebra is the first time i feel the power of the math
like as a competitor to my mind
and im really trying to lasso it in
if you know what im saying
thorn delta
you'll get used to that xd
fair shard
it also seems like such a free range playground of options
like if im proving something anywhere else
there are usually like 5 rules i can pretty much garuntee i will need
but in algebra stuff comes from all over
maiden ocean
mhm
fair shard
its beautiful and awesome to me
but its also like woah this is a beast
makes me feel so dumb
maiden ocean
its a very beautiful field i think
fair shard
cause like
my brain is still in the mode where i expect intuition to come quickly
over time that process will naturally slow down
and i will be solving problems and only after like proving them 2 different ways and using them for other things does the intuition start to form
and i guess that will have to become more normal
how long should it take before im comfortable with group theory, ring theory, and field+galois theory?
if im still working with group theory rn
carmine fossil
if $A \cong A' \text{ and } B \cong B'$ is it true $A' \rtimes B' \cong A \rtimes_{\phi} B$ for some homomorphism $\phi$ from B to Aut(A)?
cloud walrusBOT
DrunkenDrake:
chilly ocean
isn't this trivially true?
carmine fossil
The first semidirect product's homomorphism is left conjugation(k.h=khk^-1)
carmine fossil
Yes
chilly ocean
(assuming no restrictions on A,B)
carmine fossil
Basically trying to show this process generates the complete list of groups,unique upto isomorphism
next obsidian
This is gonna be really specific but
Can anyone provide an example of two integral, finite type k-algebras A,B with a map A -> B of algebras such that there exists a f.g. B-module M which is not finite over A
We can translate this to A and B being k[stuff]/prime ideal
Or maybe not even an explicit example but maybe any intuition as to what can make this fail?
I don't think the conditions on A,B are too too important but for my purposes I need to show this can fail (finiteness of M over A) when A,B are really nice rings
sturdy marsh
Do you mean finite type k-algebra?
sturdy marsh
Okay, take A = K[x], B = k[x,y], M = B
next obsidian
Oh shit
RIGHT!
Okay so the reason I asked this is
If F is a coherent sheaf on Y and you have f: X -> Y show that f_*F need not be coherent on X
even if X,Y are varieties over a field
I'm pretty sure this basically turns into finding an example of what I described above
But yeah that example is really easy to see
sturdy marsh
Does it?
In the ring case above this gives you a map from Spec B to Spec A
And a coherent sheaf on Spec B
oh pushforward
read that as pullback for some reason
next obsidian
Anyway, my idea should be correct right?
If I hit these with Spec and ~ that's my example?
sturdy marsh
yup
sturdy marsh
How's working through hartshorne going?
next obsidian
It wasn't going
since I was just
dissuaded and completely unmotivated but
It's started up again and turns out sheaves of module stuff isn't too bad
I felt really bad because the first exercise took me soooo long to do but it was because I was spending a ton of time doing all these natural isos of modules I never actually properly worked through
so I just kind of didn't do any more since I was pretty disheartened lol
sturdy marsh
lol hartshorne does that to people
next obsidian
sturdy marsh
someday I will go back and do everything again
next obsidian
🥴
maiden ocean
hurb
sturdy marsh
next obsidian
I don't know what the picture is supposed to mean haha
sturdy marsh
survivorship bias
sturdy marsh
xD
maiden ocean
does anyone agree at all on what the best AG text is
i feel like i see ppl shit on hartshorne a lot
like
next obsidian
EGA
sturdy marsh
Vakil is good
maiden ocean
"you can read all of it and not learn how to do any AG"
chilly ocean
is there an AG book for dumbasses?
next obsidian
AG book for dumbasses =
intro ones that do varieties or something I guess
or Hartshorne
since you're a dumbass for using it
chilly ocean
but i want a one that does the modern stuff
next obsidian
Then learn stacks I guess
hurb
Idk
AG is het
that's my version of calling it gay like Sham would but he's actually gay so he can do that
It'll catch on I swear
sturdy marsh
I've heard good things about mumford's red book
fierce perch
chilly ocean
@fierce perch show table of content
fierce perch
Its available online
woven obsidian
No I don't think so
I guess you can prove it by taking intersection perhaps?
Yeah that should do it
I meant for the dcc
Then we get a new prime ideal, which is finitely generated right?
Yeah it doesn't seem to work out
Ah
dim escarp
Hi, I'm redirected to here although I believe my question is nowhere near advanced
woven obsidian
It seems a bit overkill
latent anvil
you want $\sqrt[p_1]{q_1}$
woven obsidian
It's my algebraic geometry text, lecture notes written by lecturer
cloud walrusBOT
shamrock:
dim escarp
Oo thanks
How do I easily calculate degree of $Q(\sqrt[p_1]{q_1} + \sqrt[p_2]{q_2} + \sqrt[p_3]{q_3})$ over Q
cloud walrusBOT
Cascadar:
dim escarp
pi, qi's are distinct primes
We are only instructed with Galois theory and theory of cyclotomic field (in simple way)
chilly ocean
hm, what is the answer if there are only 2 terms?
latent anvil
so because this field is contained in $\Q(\sqrt[p_1]{q_1}, \sqrt[p_2]{q_2}, \sqrt[p_3]{q_3})$ we have an upper bound on the degree of $p_1 p_2 p_3$
cloud walrusBOT
shamrock:
chilly ocean
oh, but it cannot be p1, or p1p2 because reasons
cloud walrusBOT
shamrock:
latent anvil
@chilly ocean why can't it be those?
chilly ocean
i don't know, it was my bullshiting
latent anvil
lol
dim escarp
Intuition says that as well but problem is proving it without highly involved machinary
dim escarp
Because I did not learn them yet
chilly ocean
it has been too many years since i have done any algebra, these days i just bullshit things that seem true
dim escarp
Also I guess I hate this professor
He loves this hard problems
On a weekly homework
latent anvil
so weirdly my first thought is to look at the primitive element theorem
dim escarp
What is primitive element thm
latent anvil
because the proof tells you certain linear combinations of the generators of a field are a primitive element
@dim escarp it says that any finite extensions $K/\Q$ can be generated by a single element
cloud walrusBOT
shamrock:
latent anvil
I'm just thinking about how I might oslve this problem
dim escarp
$\sqrt[p1]{q1}$
cloud walrusBOT
Cascadar:
latent anvil
Because the proof actually proves that all but finitely many choices work
and im wondering if it gives a better description than that
i don't think so, but it's worth looking at
okay so
the galois conjugates of $\sqrt[p_i]{q_i}$ are $\zeta_{p_i}^k \sqrt[p_i]{q_i}$, right?
cloud walrusBOT
shamrock:
latent anvil
where k goes from 0 to pi-1
dim escarp
I mean, I only need to prove that $$\sqrt[p_1]{q_1}+\sqrt[p_2]{q_2}$$ does not generate the whole field Q(all the stuff)
latent anvil
doesn't generate which field?
dim escarp
And yes, its conjugate is that except for the range
latent anvil
oh what did I get wrong with the range?
cloud walrusBOT
Cascadar:
latent anvil
np lol
why is that element even in $\Q(\sqrt[p_1]{q_1} + \sqrt[p_2]{q_2} + \sqrt[p_3]{q_3})$?
cloud walrusBOT
shamrock:
latent anvil
or did you mean $\Q(\sqrt[p_1]{q_1}, \sqrt[p_2]{q_2}, \sqrt[p_3]{q_3})$?
cloud walrusBOT
shamrock:
latent anvil
okay, so let me see if I understand your logic
if the degree is < p1 p2 p3 then by relabeling we have $\sqrt[p_3]{q_3} \notin \Q(\sqrt[p_1]{q_1} + \sqrt[p_2]{q_2} + \sqrt[p_3]{q_3})$. Then $\Q(\sqrt[p_1]{q_1} + \sqrt[p_2]{q_2}, \sqrt[p_3]{q_3})$ contains both $\sqrt[p_3]{q_3}$ and $\sqrt[p_1]{q_1} + \sqrt[p_2]{q_2} + \sqrt[p_3]{q_3}$, so it has strictly bigger degree
cloud walrusBOT
shamrock:
latent anvil
this handles the $\deg \Q(\sqrt[p_1]{q_1} + \sqrt[p_2]{q_2} + \sqrt[p_3]{q_3}) = p_1 p_2$ case, but what if $\deg \Q(\sqrt[p_1]{q_1} + \sqrt[p_2]{q_2} + \sqrt[p_3]{q_3}) = p_1$ or $\deg \Q(\sqrt[p_1]{q_1} + \sqrt[p_2]{q_2} + \sqrt[p_3]{q_3}) = 1$?
cloud walrusBOT
shamrock:
latent anvil
why?
dim escarp
Trivial
latent anvil
why is $\sqrt[p_1]{q_1} + \sqrt[p_2]{q_2} + \sqrt[p_3]{q_3}$ irrational?
cloud walrusBOT
shamrock:
dim escarp
Oh
latent anvil
yeah lol
you cant just say trivial
I think my primitive element theorem argument actually works
but I might be wrong, I'm very tired
oh wait
dim escarp
I learned about the theorem
dim escarp
Tho what that tells me is pretty slim
latent anvil
so like, that proof gives a simple condition on what f's work
but I think the condition is still hard to verify hree
dim escarp
Basically if I can show this two:
$$\sqrt[p_2]{q_2} not in Q(\sqrt[p_1]{q_1}) and \sqrt[p_3]{q_3} not in Q(\sqrt[p_1]{q_1}, \sqrt[p_2]{q_2})$$
cloud walrusBOT
Cascadar:
latent anvil
you'd need to know that $\sin(k/p_1 \cdot 2 \pi) \sqrt[p_1]{q_1} + \sin(j/p_2 \cdot 2 \pi) \sqrt[p_2]{q_2} \neq 0$ and $\sin(k/p_1 \cdot 2 \pi) \sqrt[p_1]{q_1} + \sin(j/p_2 \cdot 2 \pi) \sqrt[p_2]{q_2} + \sin(\ell/p_3 \cdot 2 \pi) \sqrt[p_3]{q_3} \neq 0$ for all $k, j, \ell$ to apply the thing I was thinking of
which seems hard
cloud walrusBOT
shamrock:
latent anvil
I don't see why that's sufficient?
also i am going to sleep because it's 4am where I live
good night
dim escarp
Good night!
Actually I guess I was thinking in terms of Galois theory
Thought about proving degree of F(q1rp1, q2rp2, q3rp3) over F is p1p2p3
latent anvil
yeah, that is gonna be important later
And you'll do it the way you describe
I think it won't be so bad?
dim escarp
Where F is cyclotomic field
latent anvil
but first you gotta show that Q(sum) = Q(all three roots separately)
oh are we working over a cyclotomic field?
I thought we were just over Q
dim escarp
Nah, we are working over Q
latent anvil
ah okay
dim escarp
But if like
If I could show that degree of F(that one) over F is p1p2p3
I think the degree on Q follows
latent anvil
yeah, I think so too
dim escarp
With Galois field extension I could examine the subgroups of the galois group
Group of order p1p2p3 is quite nice, nice enough to say that F(a+b+c) = F(a, b, c) where a b c is you know what
So yep, if only F(a, b, c) is behaving nice
latent anvil
is it? All I know about pqr groups is that they have a normal r subgroup (where r is the largest of the three)
which is good but I don't see how it tells you anything useful here
okay I'm gonna go to sleep for real now lol
dim escarp
Okay
I mean, pqr is going to be p group * q group * r group
Each group are uniquely determined
So we have like only 8 subgroups
Btw I wish you have a good night this time =>
marsh fractal
can anyone explain in a kind of intuitive way how Euclidean Rings are Principle Ideal domains?
dim escarp
I don't think that is intuitive
sharp sonnet
i think the proof is pretty intuitive?
it's the first thing you would do and it works pretty easily
marsh fractal
ok ill look over the proof
I just dont find ERs that intuitive
the first property norm(a) < norm(ab) makes sense
but a = bm +r => r = 0 or norm(r) < norm (b) im not so sure about
dim escarp
Yep that is simply division
marsh fractal
so no remainder is the r=0 case, otherwise norm(r) must be < norm(b)
as it is the remainder
sharp sonnet
yeah
marsh fractal
ok thanks
sharp sonnet
just thing of division with remainder in integers
dim escarp
With integers you got
dim escarp
abs(r) < b
sharp sonnet
here the norm is degree
marsh fractal
I think the notation just scared me
so we have the division and multiplication structure holding on an ER
or euclidean division and multiplication i should say
so in that way ER => PID does kind of make sense
dim escarp
Oh wow
marsh fractal
the elements are increasing, even though we know nothing of the elements
dim escarp
It does not make sense to me tbh even knowing all that
sharp sonnet
the intuition is "euclidean division is powerful"
marsh fractal
the intuition is "euclidean division is powerful"
can you elaborate on this?
i drew something different
dim escarp
Sure, ED is powerful
sharp sonnet
well, euclidean rings are pretty high up in the "hierarchy of rings"
dim escarp
Too powerful that it is nearly a field
sharp sonnet
and all this structure just follows from having a suitable version of euclidean division
dim escarp
Meh my HW problem is killing me
marsh fractal
and all this structure just follows from having a suitable version of euclidean ...
can you elaborate on this more? 😅
dim escarp
They mean Euclidean domain is simply a (commutative) ring with euclidean division I think @marsh fractal
dim escarp
And it is abelian group on addition..wait
marsh fractal
ok i guess i dont have another type of division to contrast euclidean division to
as that is the only division i have used lol
but i follow what you are saying
sharp sonnet
i mean in rings you generally can't divide things
dim escarp
Being rational
dim escarp
But field is also ED..
sharp sonnet
indeed, but fields are very much more powerful than rings
dim escarp
Not powerful enough to let me prove degree of F(a, b, c)/F easily
sharp sonnet
seems like a problem on your end
dim escarp
Where a, b, c are p1, p2, p3-th roots
sharp sonnet
not on the end of the theory of fields
marsh fractal
yeah if you can divide then usually you would have multiplicative inverses
dim escarp
Prob me not being smort enough
Idk how this Galois guy got so much progress when so young
In so ancient times
marsh fractal
maybe this is a good question: why are ERs not fields?
dim escarp
@marsh fractal Well Z is ER
sharp sonnet
because you can always perform division without a remainder
marsh fractal
in a field you can define a norm that is constant 1
a norm onto the positive integers?
how are norms usually considered
a map to the positive integers with 0 ?
sharp sonnet
norms are a function into positive integers
marsh fractal
ok cool
sharp sonnet
my point is that if the norm is constant 1
then in euclidean division the remainder will always have to be 0
so you can divide any 2 elements
that are nonzero
or in other words
every nonzero element is a unit
marsh fractal
yes
sharp sonnet
i.e. your ring is a field
marsh fractal
because multiplicative inverses exist correct?
so you can define a norm that assigns everything to 1
norm = aa*
a* being a's inverse
dim escarp
Norm isn't necessarily aa*
marsh fractal
but is it not always a possible norm to make
for fields
do norms always exclude the 0 element?
sharp sonnet
ye
dim escarp
In this context
marsh fractal
ok
sharp sonnet
for field you can have a norm of constant 1 always
dim escarp
Idk why it called norm
sharp sonnet
it will be valid
dim escarp
Another aboose of notation?
sharp sonnet
ah, its unrelated to norms in analysis
marsh fractal
yeah i am seeing now haha, it is its own thing in algebra
but ok thanks that helps a lot
dim escarp
Meh I guess I'll just let this problem as first problem I failed to solve
marsh fractal
nope you helped!
dim escarp
@marsh fractal oh I was not talking abt you
marsh fractal
ok lol 🙂
dim escarp
I was talking about the HW problem I forementioned
marsh fractal
didnt see that one
dim escarp
I did not know getting degree of F(a,b,c)/F is so hard
Simply matter of Degree of extension, but hard
marsh fractal
yeah i am not there yet unfortunately.. I would like to help
dim escarp
woven obsidian
I'm not sure how they conclude the thing at the bottom. That f maps surjectively into W_1
next obsidian
I’m not gonna lie, I don’t know shit about varieties.
It sounds like maybe this is going up though, which follows since f is finite?
woven obsidian
I think the going up part is used first, for the second part I'm not sure what they do
It wouldn't surprise me if my lecturer swept tons of details under the rug
next obsidian
I mean it sounds like Q is a prime?
And the existence of a prime such that... sounds like some sort of going up kind of deal but because it’s in variety language I have no clue what’s going on ;w;
Or that P is... i don’t really know 😅
Tfw can’t do varieties
chilly canyon
Hi ! I'm having trouble visualising the Extension of a group... I'm needing this in the Universal Coefficient theorem in cohomology, but that's not the problem :)
If I take $A=\mathbb{Z}^k\oplus Tor$ the free-abelian/torsion decomposition of $A$, then what would be $Ext(A,\mathbb{Z})$ or $Ext(A,\mathbb{Z}/2)$ for instance ? (assuming for the latter that the torsion is a direct sum of $\mathbb{Z}/2$ for simplicity)
cloud walrusBOT
Matplotlib:
chilly canyon
There are no examples on the Wikipedia page...
maiden ocean
I'm not sure what you mean by visualizing but Ext factors through direct sums
chilly canyon
Well, I don't really know how to compute it ^^"
maiden ocean
Z^k is free so Ext(Z^k) = 0 and thus this is just Ext_Z^1(T_A)
chilly canyon
Because I'm unsure what the construction is doing
chilly canyon
Yup
maiden ocean
so like
you know how if we have a short exact sequence 0 -> A -> B -> C -> 0
and then a left exact functor from the category these are in to some other category
we get like 0 -> F(A) -> F(B) -> F(C)
(covariant functor)
chilly canyon
The last arow F(C)→0 is not there because only left exact no ?
maiden ocean
yea sorry
chilly canyon
Okay
maiden ocean
thats what makes this importnat lol
so like
the free resolution stuff is basically a way of canonically extending this into a long exact sequence
given the functor F
chilly canyon
Oh okay it's the F(C)→Ext→0 ?
chilly canyon
Yeah
maiden ocean
ya
chilly canyon
So it's the right derived of the contravariant Hom ?
maiden ocean
mhm
chilly canyon
Not sure about left/right now, but I think I get it
maiden ocean
so like it basically measures like
ok if we have 0 -> A -> B -> C -> 0
and we apply a covariant functor to get 0 -> F(A) -> F(B) -> F(C)
the derived functor basically measures how far this sequence is from being exact
maiden ocean
yea
chilly canyon
Okay !
maiden ocean
and like
chilly canyon
And we do the same but the opposite way for a contravariant with G(C)→G(B)→G(A)→0
chilly canyon
Yes
maiden ocean
and we identified Ker i_n* with Hom(H_n(C), G)
so the reason coker identifies with the Ext is bc any abelian group has a free resolution 0 -> F_1 -> F_0 -> H -> 0
so the Ext stuff is all trivial for i > 1
next obsidian
Cuz Z is a PID 
chilly canyon
Okay I think I get it
maiden ocean
meaning that Ext^1 measures how much ur sequence fails to be exact and can be identified with Coker
next obsidian
Moth knows cohomology
maiden ocean
next obsidian
Is this from AT?
maiden ocean
yes
chilly canyon
Yeah !
chilly canyon
Thank you Moth ^^
uncut girder
Tor is the hidden structure behind tensor product
next obsidian
I just like Tor because so many useful things can be phrased in terms of clever tensor products
And you can calculate torsion stuff with it, blah blah
So having Tor is really nice
maiden ocean
Kunneth formulas go brrrr
maiden ocean
Smh literally all of cohomology
next obsidian
🥱
I’ll do homological algebra next quarter smh
Also this meme brought to you by I know 0 AT
next obsidian
...
Moth
How many times I gotta tell ya ass
You can’t ask me this now
I’m not spiritually ready
maiden ocean
: (
next obsidian
Maybe in a years time I will be sufficiently something-pilled
maiden ocean
next obsidian
maiden ocean
next obsidian
🤬🤬🤬🤬
golden pasture
Read lurie with me
mememememe
maiden ocean
Kinda
golden pasture
algebra ug lel
next obsidian
Like your hw was like
maiden ocean
Honestly yes
golden pasture
next obsidian
Problem 2: immediate result from the fundamental theorem of Galois theory
next obsidian
Oof
Just
Look at the fundamental theorems statement over and over
So you don’t mix up which extensions are Galois and shit
Lol
maiden ocean
I will find the list of topics later today smh
next obsidian
It’s like
maiden ocean
I have bad intuition for fields
next obsidian
If F < K < L
And L is Galois over F
Which one was Galois again?
I mean I know the answer
But that stuff always makes me have to think for a second lol
And the stuff about composite extensions
next obsidian
🥱
maiden ocean
O
golden pasture
like how you know the chinese shows
ok maybe not im too boomer lmao
chilly ocean
@maiden ocean prove primitive element theorem
golden pasture
it's so sad lol i spent most of my childhood time with my grandparents so i effectively am of their style
artin primitive element theorem owo
chilly ocean
are you a boomer
golden pasture
theres a relatively simple proof for char 0 compared to the general case iirc
maiden ocean
why would i do that frucht
chilly ocean
because apparently it gives intuition
maiden ocean
hurb
chilly ocean
what is hurb
maiden ocean
hurb is hurb
golden pasture
artin primitive theorem what intuit
maiden ocean
the hurb zahlenburger dan trifecta
i am on chapter 3 of hatcher
but less than halfway through
hurb
chilly ocean
artin primitive theorem what intuit
golden pasture
i gained no new intuition learning it
chilly ocean
did you prove it yourself? @golden pasture
golden pasture
it's the kind of theorem that is like
it is morally true
it is clearly true for char 0
it is a pita for char p
cuz nonseparable pain
chilly ocean
wait why did i ping this person
chilly ocean
learn nt
golden pasture
nt is pretty cute
why are there so many things to learn
and so little time
;-;
maiden ocean
golden pasture
HAHAHAHAHA
i kinda want to learn like p adic stuff
soz like
reading neukirch chap 2&3
maiden ocean
ari have u seen the "a single good book can change your life" meme
golden pasture
uh nop
maiden ocean
its this
ok imagine the book is neukirch the men b4 are ur current pfp the ppl after is zophs pfp
golden pasture
sksksksksksksksk
maiden ocean
if i could photoshop i would make this
maiden ocean
what will u learn
maiden ocean
golden pasture
but i only like
maiden ocean
maybe we can read together
golden pasture
rudin is trivially easy to scan through
maiden ocean
pepelaugh
golden pasture
real analysis is just either chain of inequalities or constructing the most cursed example you can think off
carmine fossil
Are you a prodigy,cat?
golden pasture
idts
maiden ocean
idk what that word means
golden pasture
can you really call someone just like 1/2 years faster than usual a prodigy
prob not
maiden ocean
ppl r so obsessed with sorting everyone who does math into two types of people
maiden ocean
like either ur "normal"
or ur "a prodigy"
and these r treated like two different kinds of human being
golden pasture
it's kinda super lame tbh
maiden ocean
its like
either ur a genius who will change mathematics or it doesnt matter
which is pretty 
maiden ocean
sad!
golden pasture
thonk
maiden ocean
me
golden pasture
what is your emote
maiden ocean
maiden ocean
because the original gabe is gone
i will read it in n months
ari too
rn i am doing hatcher tho
and i still have to do rudin and AM
maiden ocean
noooooo
golden pasture
maiden ocean
ariana i stumbled on a wild negi post on MSE the other day
it was a comment
he downvoted someone for liking haskell
chilly ocean
and these r treated like two different kinds of human being
math is getting sorted into people who like comp math and people who don't
maiden ocean
hurb
chilly ocean
ariana i stumbled on a wild negi post on MSE the other day
what is negi's name on mse?
chilly ocean
alright
golden pasture
he downvoted someone for liking haskell <:egg_hank:690775538205458452>
i would tbh xd
maiden ocean
it was really funny
someone was like "why is this downvoted?" and he commented "because i dont like haskell"
chilly ocean
sad
golden pasture
next obsidian
So some AG thing has me suspecting the following is true, but I don't see how I'd prove it. I have a few very vague ideas of where to go but none of them seem like they'd pan out very well. Anyway the statement is
Let $A$ be a Noetherian ring and $M$ any $A$-module. Let $m \in M$, then if $I$ is an ideal of $A$ such that $V(\text{ann} m)\subseteq V(I)$ then there exists some $n > 0$ for which $I^n \subseteq \text{ann} m$. If ann $m$ is radical (which it totally isn't in general) I can see how to show this, but I don't really know how to proceed.
cloud walrusBOT
Chmonkey:
next obsidian
I know $I$ is finitely generated and I thought you might peer into some localizations, clear denominators yadda yadda, but this isn't quite right, and I was playing around with trying to do some bullshit primary decomposition stuff but that doesn't seem very useful either
cloud walrusBOT
Chmonkey:
next obsidian
I've already managed to prove the other direction of this, and I suspect it's an if and only if
carmine fossil
Bookmark AA
maiden ocean
hurb
queen vine
If $V(\text{Ann} m)\subset V(I)$ then $\sqrt{I}\subset \sqrt{\text{Ann} m}$
So $I^n\subset\text{Ann} m$ for some $n$
next obsidian
I think I proved it
Yeah I was looking at that
And couldn’t figure out how to proceed but using judicious use of finite generation and taking powers I think I got it
next obsidian
So $I^n\subset\text{Ann} m$ for some $n$
Lol this is what I was trying to show haha, I couldn't see how to prove this from the earlier statement
This definitely isn't immediate, I'm pretty sure you crucially need I to be finitely generated in order to show it
queen vine
It is just pigeonhole principle
this type of arguing appears very often in these problems
next obsidian
Right, but my point is just that I couldn't figure out how to do that so that response didn't really help much
I already knew that statement about the radicals but just saying the other follows is pretty much just restating what it was I wanted to prove haha
queen vine
Yeah, sorry, you said you figured out the case where ann m is radical but my point was that since $\sqrt{I}\subset\sqrt{ann~ m}$ we can assume that ann m is radical
cloud walrusBOT
leoli1:
next obsidian
Yeah, it doesn't quite work that well
Either way, I manged to figure it out since I is a subset of sqrt{ann m}
Basically doing a fat pigeon-hole thing like you said
maiden ocean
for any abelian group H ... -> 0 -> 0 -> H -> 0 is a free resolution right?
but it must not be chain homotopic to any other free resolution of H
or uh at least
it cant always induce isomorphisms on the Ext
oblique river
@maiden ocean I think resolutions need to be exact
maiden ocean
maybe im an idiot but since both maps r the zero map are they not
wait
hahahaha
never mind
oblique river
:^)
maiden ocean
sometimes i forgot that exactness is stronger than im subset ker
oblique river
yep
obsidian path
I understand that the polynomials of degree 2 and 3 in Z2[X] are only irreducible if they have no roots in Z2, so I attempted a root test f(0) /= 0, and f(1) /= 0
is that right?
maiden ocean
I mean it works
next obsidian
That’s the way I’d do it
You only have two choices for each coefficient and only two inputs
obsidian path
yes its small so its not hard to find by trial and error, but I'm trying to be rigorous
next obsidian
That is rigorous though
maiden ocean
Just list and then compute
maiden ocean
Rigor x 1000
maiden ocean
Hello slim and chmonkey
obsidian path
yeah that's what i was thinking right but i prob was overthinking it
same
analysis is where its at
topology baby
not that ive studied it that much yet but
yeah I feel that in algebra a lot and thats ok
maiden ocean
Topology without algebra would be lame
AT is great
Well ok i havent done diff top or whatever
TTerra moment
latent anvil
does the tensor hom adjunction preserves isomorphisms?
Like if I have an iso f : L (×) M -> N, is the corresponding map L -> Hom(M, N) an iso?
no, L = Z/2Z, M = Z/3Z, N = 0 (working over Z)
dim escarp
Let me try again
How do I easily calculate degree of $Q(\sqrt[p_1]{q_1} + \sqrt[p_2]{q_2} + \sqrt[p_3]{q_3})$ over Q when $p_i q_i$ are distinct primes
cloud walrusBOT
Cascadar:
dim escarp
I tried using cyclotomic field where I could only need to show like
Degree of $F(\sqrt[p_1]{q_1}, \sqrt[p_2]{q_2}, \sqrt[p_3]{q_3})$ over F
cloud walrusBOT
Cascadar:
golden pasture
it is a subfield of $\mathbb Q\left(\sqrt[p_1]{q_1}, \sqrt[p_2]{q_2}, \sqrt[p_3]{q_3}\right)$ so what does this tell you about its possible degrees
cloud walrusBOT
ariana:
dim escarp
I do know it's divisor of p1p2p3
Problem is
How do I prove the equality when finding membership is too uard
golden pasture
theres prob a simpler solution but we know the extension Q(pth roots)/Q is galois and the subgroup of this that fixes Q(sum) is trivial i think
latent anvil
theres prob a simpler solution but we know the extension Q(pth roots)/Q is galoi...
If you just adjoin the real pth roots you don't get a Galois extension
and if you adjoin all the roots the subgroup which fixes Q(sum) will be bigger
Yesterday Cascadar suggested we look at $F = Q(\zeta_{p_1},\zeta_{p_2},\zeta_{p_3})$ and consider $F(\sqrt[p_1]{q_1}, \sqrt[p_2]{q_2}+ \sqrt[p_3]{q_3})$ as a subfield of $F(\sqrt[p_1]{q_1}, \sqrt[p_2]{q_2}, \sqrt[p_3]{q_3})$
cloud walrusBOT
shamrock:
latent anvil
which made sense to me
latent anvil
so I guess first you'd want to say $\sqrt[p_1]{q_1} \notin F, \sqrt[p_2]{q_2} \notin F(\sqrt[p_1]{q_1}), \sqrt[p_3]{q_3} \notin F(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2})$
Which seems annoying
cloud walrusBOT
shamrock:
dim escarp
@golden pasture wdym pth roots
Oh you gotta sleep
It's enough to show $\sqrt[p_i]{q_i} \notin F$ because of degree thing
cloud walrusBOT
Cascadar:
dim escarp
Notably, degree of $F(\sqrt[p_i]{q_i}, others) : F(\sqrt[p_i]{q_i})$
cloud walrusBOT
Cascadar:
dim escarp
<@&286206848099549185> anyone got a clue
I'm not able to leverage full power of field theory yet as am in introductory course
dim escarp
Lol I found the answer myself
Simply not being Galois itself makes it not contained
Lmaooooooooooooooooo
Lol I'm mad now
\kill me
uncut girder
@dim escarp it's more like the galois closure of Q(cube root of 2) is nonabelian, where as every cyclotomic extention is abelian. Thus no cyclotomic extension can contain Q(cube root of 2)
dim escarp
@uncut girder I mean, that sounds more like going around
In the answer, it explains that any subgroup of cyclotomic extension is normal becuz abelian
uncut girder
No, the nonabelian thing is the most important part
Any abelian extension will be contained in a cyclotomic field
dim escarp
Well for that you have to prove that the galois closure is nonabelian
Which isn't as easy I think
uncut girder
I think it's the simplest example of a nonabelian extension
dim escarp
I mean it's simpler to say Q(cube root of 2) is not Galois over Q
Why would you induce extra steps
"Any subextension of cyclotomic extension is Galois" gives simpler proof
While having to describe Galois group of Galois closure involves much more step
valid scaffold
Hi how can solve: Use SVD to show the push-throughany n × d matrix D and scalar λ > 0:
uncut girder
@dim escarp yeah it's just my natural way of thinking about it
But you're right
For this application that's easier
dim escarp
Also easier for my problem
valid scaffold
<@&286206848099549185>
chilly ocean
Using svd? I think it suffices to show that the equation holds where you clear the inverse matrices
Like instead of A^-1 B = C you make it B = AC
valid scaffold
i dont think i follow you, so i can pass the first term to the right and just leave D^T?
chilly ocean
Yeah exactly, and you can do the same for the inverse matrix on the right side
As long as you are assuming that lambda * I + D^T D is invertible in the first place (and same for D D^T)
valid scaffold
So it is like they exchange?
chilly ocean
exchange?
i'm just saying you can easily show $$D^T (\lambda I_n + DD^T) = (\lambda I_d + D^TD)D^T$$
cloud walrusBOT
88ddda:
valid scaffold
Oh ok yeah, that's what I got from your explanation, I just thought there would be more steps and thanks
and sorry for the incorrect use of exchange, I think I did not express myself well.
dim escarp
I didn't know q-th root of p not being in cyclotomic was such an easy matter
fair shard
wtf is fourier analysis on groups
golden pasture
relatable
fair shard
the only fourier I know is like... linear are
algebra*
is it at all related?
I can't possibly imagine how it could be
fourier must have been doing some side math
golden pasture
from my ε knowledge on the existence of this topic it's some representation theory thing prob have related ideas
in physics we do have like e^iHt stuff do ig it is somewhat related
fair shard
I hate physics 😍
golden pasture
is this a sign of high school trauma
fair shard
no
just don't care for applications...
well actually
I know barely anything about physics
but....
I think it's the introductory stuff that bores me
I never ever want to calculate the work done over a surface or wtf like ide
ide
idek*
golden pasture
lol same
fair shard
but quantam mechanics and relativity look pretty cool :)
golden pasture
i think generally it's the more high school lvl/intro ug physics that have such computations tho
fair shard
I bet when I learn differential geometry I will care for relativity
golden pasture
fair shard
I heard there's something about curvey things
anywaysssss, discrete>continuous is my motto
pure > applied
both of those are broken by physics, so I never loved it
golden pasture
i like landau sr+em treatment but his gr kinda sucks lol he doesnt rlly use diff geo until gr so if you're interested can check landau book 2 out
maybe you would like stochastic memes in physics🤔
fair shard
wtf is that like random variables and stuff flying all around the room?
bleak abyss
DID SOMEONE SAY FOURIER ANALYSIS ON GROUPS
latent anvil
@fair shard doing diff geo now
do not care about relativity
still hate physics
also @bleak abyss you should give ttera honrable
next obsidian
Hey Dami
latent anvil
and i want to learn fourier analysis on groups at some point
bleak abyss
Yeah how've you been chmonkey?
latent anvil
chad
bleak abyss
Damn
crude sail
quick question
calculus would prolly be considered the less rigorous, more computational version of real analysis
is there an analogue to calc but for abstract algebra
thorn delta
a non-rigorous intro to linear algebra maybe
golden pasture
ye
crude sail
cool cool
golden pasture
computational would prob be like trying to implement things just follow the algorithm given
chilly ocean
maybe playing with a rubiks cube
golden pasture
ok now that i am not that sleep deprived
Let
$$K=\mbb Q\left(\sqrt[p_1]{q_1}+\sqrt[p_2]{q_2}+\sqrt[p_3]{q_3}\right)$$
and let
$$L=\mbb Q\left(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2},\sqrt[p_3]{q_3}\right)$$
Let $$M=L\left(\zeta_{p_1},\zeta_{p_2},\zeta_{p_3}\right)$$ be the normal closure of $L/\mbb Q$ with elements in the galois group determined by where it sends $\zeta_{p_1},\zeta_{p_2},\zeta_{p_3},\sqrt[p_1]{q_1},\sqrt[p_2]{q_2},\sqrt[p_3]{q_3}$. Let $H=Gal(M/L)$. We want to show $H=Gal(M/K)$ as well. Assume there exists some $g\in Gal(M/K)$ and $g\notin H$, then g must permute say $\sqrt[p_1]{q_1}$ to $\zeta_{p_1}^k\sqrt[p_1]{q_1}$, but this cannot fix $K$ cuz it also changes the generator of $K$, hence $K=L$
There may be a simpler solution by considering the case of $\mbb Q\left(\sqrt[p_1]{q_1}+\sqrt[p_2]{q_2}\right)=\mbb Q\left(\sqrt[p_1]{q_1},\sqrt[p_2]{q_2}\right)$ first cuz along the lines of $x^{p_2}-q_2$ has no root in $\mbb Q\left(\sqrt[p_1]{q_1}\right)$ and then we must have $$\mbb Q\left(\sqrt[p_i]{q_i}+\sqrt[p_j]{q_j}\right)\subset\mbb Q\left(\sqrt[p_1]{q_1}+\sqrt[p_2]{q_2}+\sqrt[p_3]{q_3}\right)$$ for distinct $i,j$, hence the degree must be divisible by $p_ip_j$ and it is only possible if it is $p_1p_2p_3$
cloud walrusBOT
ariana:
golden pasture
How do I easily calculate degree of $Q(\sqrt[p_1]{q_1} + \sqrt[p_2]{q_2} + \sqrt...
(in relation to this)
dim escarp
Thanks, but I solved it in another way. Btw.. why g must permute $\sqrt[p_1]{q_1}$
cloud walrusBOT
Cascadar:
dim escarp
When g is not in Galois group of M/K
golden pasture
cuz elements in Gal(M/K) are those that fixes $\sqrt[p_i]{q_i}$
cloud walrusBOT
ariana:
dim escarp
Hmm, I mean I see no reason it should send $\sqrt[p_i]{q_i}$ to $\zeta \sqrt[p_i]{q_i}$
cloud walrusBOT
Cascadar:
dim escarp
For me it looks like it might be able to send $\sqrt[p_1]{q_1}$ to other roots like $\sqrt[p_2]{q_2}$
cloud walrusBOT
Cascadar:
dim escarp
Hm I guess that part I could work out, but then, how do you know that it can't fix generator of K?
golden pasture
For me it looks like it might be able to send $\sqrt[p_1]{q_1}$ to other roots l...
ah but then $\sigma\left(\sqrt[p_1]{q_1}\right)^{p_1}=\sqrt[p_2]{q_2^{p_1}}$ but $\sigma\left(\sqrt[p_1]{q_1}\right)^{p_1}=\sigma\left(q_1\right)=q_1$ cuz $q_1\in\mbb Q$
cloud walrusBOT
ariana:
dim escarp
Ya I think that one can be worked out easily
Tho why it does not fix generator of K
The na+mb+lc != a+b+c kind of thing does not look so obvious to me
dim escarp
It certainly doesn't seem trivial
golden pasture
theres the nice reason cuz this field is purely real -> the real parts of sum of any conjugates will be less
cuz the real part of ζ_p^k q^(1/p) is strictly less than q^(1/p)
dim escarp
Oh, interesting
carmine fossil
What is the most efficient way to show the only non abelian groups of order 8 are Q_8 and D_8?(and order n in general)
next obsidian
Lol order n in general isn’t known
If you knew that you’d have a classification f all finite groups
Anyway the only way I know how is to just classify groups of order 8 using semi direct products.
carmine fossil
Q_8 won't be a semidirect product,tho
next obsidian
This is to show if you aren’t Q8 you have to be D_8
You know the number of elements of order 2 must be odd so it’s either 1,3,5,7 if it’s 7 then you have to be C2^3
I believe if you assume there’s only 1 you can show it’s Q_8, I forget how
Then if you’re nonabelian but have more than 1 element of order 2 you can do a semi direct product of C2 and C4 or something to get D_8? I forget precisely
carmine fossil
C_4 semi C_2 works
next obsidian
I don’t recall the specifics, but really I guess my point is just that you have to classify all groups of order 8
Since well... if you classify non abelian ones of that order you get them all
And I think you want to look at elements of order 2 to do that
It’s been a while since I did it though
So if you have more than 1 element of order 2 you have at least 3. You need at least one element of order 4
So we can get a copy of C4 and at least 3 C2’s
There’s only 2 elements of order 2 in that C4 so one of the C2’s and the C4 are trivial intersection
So you get C4 semi C2 which is always D_8 if it isn’t just C4 x C2?
So you have to handle the case of identity, one element of order 2, and 6 of order 4
I forget how you show this has to be Q8 lol
Oh right so for the C4 semi C2 the normal@one is C4
So there’s only two possible maps you could define it by I think
This shows it’s just C4 x C2 or D8 if you can show D8 is some C4 semi C2
Which you can do
Hmm how to handle the case of one element of order 2 tho
carmine fossil
If there is no element of 8,there has to be 3 subgroups of order 4(because 2 distinct subgroups always share 1 and the element of order 2)
next obsidian
Right
carmine fossil
Let's the say the three subgroups are generated by x,y and z. Now xy cannot be in <x> or <y> ,which implies xy=z(or z^-1)
next obsidian
Right
So do you think you can just arbitrarily map i,j,k to these?
z or z^-1 as needed
I think so maybe
So because of those relations this should make a well-defined map
Since the relations on the end preserve the ones in Q8
carmine fossil
I think so
next obsidian
Shit I need to look up a presentation of Q8 haha
carmine fossil
Map i->x and j->y and you are done
next obsidian
I always forget what’s the least you can get away with
So we know that x^2, y^2, z^2 are the element of order 2?
carmine fossil
Because (x^2)^2=x^4