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00:14
I have already found a counterexample guys, thanks.
 
2 hours later…
02:35
0
Q: Size of fibers versus degrees of residue fields in a morphism of affine schemes

Soham ChowdhuryConsider $\DeclareMathOperator{\Spec}{Spec} \newcommand{\F}{\Bbb F} \DeclareMathOperator{\Z}{\Bbb Z} \Spec \Z[i]\to\Spec\Z$ given by the inclusion of $\Z$ into $\Z[i]$. Now, the fiber over every prime which is 1 mod 4 has two points, e.g. $f^{-1}((5)) = \{(1+2i),(1-2i)\}$. Over primes congruent ...

@Balarka, I asked on main.
 
1 hour later…
user227867
03:49
@Alessandro I will try to see 'Children of men', thanks.
user227867
@amWhy I got your latest email. I don't know what you are trying to say, but never mind.
user227867
05:57
@amWhy No, I have not.
06:16
@SohamChowdhury Doesn't seem to have any geometric motivation behind inert primes there in the answer though.
user227867
@BalarkaSen Have you decided which university to go to for your studies?
user227867
I can't believe I got 2 stars for that stupid comment, lol.
It's a comment I also agree with.
user227867
I like some Indian authors, like V S Sunder, R Narasimhan, M Ramachandran, and S Varadarajan.
user227867
06:24
The problem is that many of their names are too long for me to remember. =)
Indeed.
user227867
At least B Sen is easy. =)
"BS" in short.
user227867
Like Brian Scott.
I had a different de-concatenation in mind.
Hi @Huy
user227867
06:30
Like Bull Shit.
Huy
Huy
hey @Balarka
@JasperLoy Exactly
user227867
I think math degrees should not be called Bachelor of Science but of Arts instead, since math is not a science.
@Huy What are you working on?
Huy
Huy
@BalarkaSen: reseasoning my cast iron skillet. just woke up, having coffee, afterwards brunch at my parents' place
and in the afternoon, I'll prepare some more geometry for my students, then probably take a break
so mathematics-wise, not much at the moment, sorry to disappoint
maybe I'll study a bit about quaternions to see if there's something interesting that I can show the kid that I didn't know yet
user227867
06:33
@Huy Also mention octonions.
08:07
hi
08:29
hiiiiiiiiiiii
user227867
Hi @usukidoll, lol
can someone look over one of my proofs? I think I got them but the way I worded it is crap
user227867
If the way you worded it is wrong, then maybe the proof is wrong.
I got this
If $a \vert b$ and $c \vert d$, prove that $ac \vert bd$\\
By the Divisibility definition, a divides b if $b=ax$ for some integer x and c divides d if $d=cy$ for some integer y. Multiplying b and d, we have, \\
$(ax)(cy)=bd$\\
$(ac)(xy)=bd$\\
Hence, $ ac \vert bd$\\
user227867
Anyway, I am only a banana.
user227867
08:34
@usukidoll You mean ac in the second line.
user227867
ac divides bd in the second line you typed
oh shit typo
user227867
Yeah, shit, lol
now it sounds legit
user227867
08:36
@usukidoll Just mention that xy is integer. That is it.
like since xy is an integer, multiplying b and d results in
(ax)(cy) = bd
(ac)(xy) = bd
Hence ac divides bd?
user227867
@usukidoll You should mention it just before the conclusion.
ef
errr
UGH!
user227867
@usukidoll I will show you how I would write it.
user227867
08:41
@usukidoll Since $a$ divides $b$, $b=ax$ for some integer $x$. Since $c$ divides $d$, $d=cy$ for some integer $y$. Now $bd=(ax)(cy)=(ac)(xy)$ and $xy$ is an integer. Hence $ac$ divides $bd$.
user227867
Is that alright @usukidoll?
user227867
Do you think you should thank me now?
yeah gives chocolate frog
 
1 hour later…
10:13
Hey
r9m
r9m
I am not particularly good with these stuff ... a friend asked :) so asking for help here: math.stackexchange.com/questions/1906134/… thanks! :)
hey @usukidoll .. I see you are a weird frog avatar! :P
I think it's a Pokémon.
bulbasuar
bulbasaur
u there?
10:30
hey
fellow numberfriends
What if you only like topological spaces?
what's the gcd(3,5)?
really?! well I was dividing a lot and got 1
but when I did the first part I got 2
Isn't it quite obvious? They're both primes.
10:34
like 3 divided by 5 but the remainder was 2...
then I had gcd(2,3) did this again and got a remainder 1
well I was using EUclidean ALgorithm
Show by example that if $au+bv=d>1$, then $(a,b)$ may not be d.\\
I let a =3, u=4, b =5, v = 6
(3)(4)+(5)(6)=12+30=42
but gcd(3,5) is 1
And?
that's it?!
42 is greater than 1 but I highly doubt that gcd(3,5) produces 42
What do you mean with (a, b)?
10:39
my d turned out to be 42 when I let a =3, u = 4, b =5 v = 6
But what do you mean with the notation (a, b)?
but gcd(3,5) is 1 so (a,b) isn't d which was 42.. right?!
like gcd(a,b)
it's the notation from abstract algebra an introduction 3rd edition
they left out the gcd
Not everyone is familiar with the notation used in your book. It's good to be unambiguous.
But doesn't Bezout's theorem contradict what you are saying?
O_O!
so am I wrong?
Well take $a = 2, \; b = 4, \; u = -1, \; v = 1, \; d = 2$.
$$(2)(-1) + (4)(1) = gcd(2, 4) = 2$$
10:58
Show by example that if $au+bv=d>1$, then $(a,b)$ may not be d.
MAY NOT BE D THOUGH!!!!
so I'm proving that by example
Ooh, misread that, my mistake.
my D was 42 but when gcd(3,5)= 1
$ 1 \neq 42$
You can easily construct a solution that fulfils your condition.
a=3,u=4,b=5,v=6
12+30=42 but gcd(3,5) = 1 whoops not equal
Let $a, b \in \mathbb{Z}$ and let $u, v$ be integers such that Bezout's relation $au + bv = gcd(a,b)$ is fulfilled. Let $d = 2 \times gcd(a,b)$ then $$2au + 2bv = 2 \times gcd(a,b) = d$$ Then your solution is the quadruple $(a, \; b, \; 2u, \; 2v)$.
11:04
but I'm wondering if what I did was right?! since 42 is d but the gcd(3,5) was really not d
It is clear that $d > 1$.
Yes, it's correct.
yay ^_______^
11:18
the sum of three squares
$n \equiv 3(mod4)$
I think $3^2+3^2+3^2=27$ which is 3 in mod 4
@Huy Ah, I see.
Busy with teaching, I see.
Huy
Huy
quite a bit, mostly because I've never taught 9th graders before, so I need to get some experience how much will be enough for how long
I kind of underestimated them at first, so now I'm trying to adapt
I have this strange experience in here that a generic 9th grader is more mature (in the sense of being more serious about learning, actively thinking more and asking questions, quick at understanding, etc) than a generic 11th grader.
Huy
Huy
11:34
no, it's completely the opposite here
Might be a coincidence because I have only actively interacted with 3 batches (one being my own, when I was a 9th graders, the next one, and the one now - I am in 11th now)
Huy
Huy
mostly because the school where I work at is for 9th till 12th grade, so 9th graders are new and very self-confident and believe they are the smartest kids alive, but after 1-2 years at the school, they realise how much they still have to learn and become much more humble
Huy
Huy
and since most teachers just do revision at the beginning to make sure everyone's on the same level, they are convinced they know everything already
Yikes, yes, dealing with overly excessive self-coincidence is a bit of a trouble.
Huy
Huy
11:36
yes, I'm glad I don't only teach 9th grade this year
Is the quaternion guy from 9th?
Huy
Huy
yes
I have a bad feeling about his eagerness to learn quaternions then
Huy
Huy
same
I'll find out soon enough
pro-tip: just don't be the guy who's teaching us physics and says eagles fly at escape velocity
just a random suggestion. might be useful
Huy
Huy
11:43
I'm not teaching them any physics any time soon anyways
I need a light bulb. Can't solve the problem I want to.
Huy
Huy
buy Philips Hue. they make you smarter.
you should send an ad-proposal to Philips on that theme.
user227867
12:10
@Huy You should not teach quaternions to a 9th grader. If he is so inclined, he can learn it himself, lol.
@JasperLoy Why not?
If someone is genuinely interested, one should not discourage - rather encourage - based on intellectual maturity, IMO.
user227867
@BalarkaSen If a 9th grader is talented enough to learn quaternions, he can learn it himself. If he isn't talented enough, then he has no business learning it.
I do not agree with that point of view. One doesn't need to be talented to be interested in learning something.
user227867
If one cannot understand 1+1=2, he should not try to understand 2+3=5.
user227867
I feel that too many people are trying to learn things beyond them these days.
12:20
Maybe he does understand complex numbers and so on. Maybe he does understand the definition of quaternions but wants to know what it's good for.
I don't think one should deny outright to help someone learn something beyond the textbooks. This is the kind of attitude which discourages most people to learn anything at all, especially in high school.
user227867
Then he can learn it himself, like I said.
No, I don't think this is true.
user227867
For me, I learnt everything on my own.
E.g., I am interested in learning quantum mechanics. I probably can't learn it myself without significant guidance.
user227867
Another problem is finding the wrong guidance.
user227867
12:25
@BalarkaSen I think you can. You are genius.
I tried to put some effort in understanding the basics. I don't understand everything satisfactorily well enough in the little of the book I have read.
user227867
Yeah, we need to read a few books sometimes to see things. One needs a good book, and one needs a good teacher.
user227867
Not all books are good, and likewise not all teachers are good.
Yes, one needs a good teacher. A book doesn't explain how to think - a teacher does.
user227867
Some people learn to think themselves.
user227867
12:30
I think we don't disagree at all. Maybe I just feel that it is too young for someone to know about quaternions.
user227867
Maybe I have seen too many cranks trying to learn that 2+3=5 when they don't know 1+1=2.
Huy
Huy
@BalarkaSen I already work for them.
should have known
Huy
Huy
I'm an excellent advertiser.
user227867
13:14
@usukidoll If you know the meaning of gcd it is clear that it is 1.
13:34
@r9m interesting
will switch to this when i m done working
r9m
r9m
14:16
@Agawa001 indeed :) I was stumped with this problem ..
14:36
@Huy quaternion guy?
Huy
Huy
?
who is the quaternion guy
Huy
Huy
kid who wants to learn about quaternions. pretty sure you've been here when we discussed it
maybe but I was probably doing this god damn analysis problem set
@r9m have u tried graphs ?
Huy
Huy
14:38
lol
@Huy is this the guy you were wondering about what to teach
Huy
Huy
y
r9m
r9m
@Agawa001 I ain't good with graphs .. :|
@Huy did he say why?
I find them totally boring
hey im given that my matrix is 3x4 and the nullity is 2 so the rank of my matrix is 2 how do i find a basis for the row space?
im not given any matrix
Huy
Huy
14:47
@0celo7 that was pretty much exactly my thought
only the null space
Huy
Huy
my theory was: probably just "other kids only know rational/real numbers, some smart kids know complex numbers so I must know quaternions !!"
i kno that the row space has dimension 2 subsdet fo $R^4$
@Huy him not knowing calculus is crappy
Huy
Huy
idd
14:49
idd?
Huy
Huy
indeed
@r9m me niether but being not good is better than not experiencing it at all
i have long history with graphs
the biggest graph structure i mad was the air of my room wall
@Huy how are the areas of spheres related to the volumes of the balls they bound
Huy
Huy
are you asking me or telling me to teach him
asking
there is a neat physicist proof of the volume of an $n$-sphere
but it uses calculus
user227867
14:52
@user1618033 Hi!
Huy
Huy
yes
is there a relationship without calculus?
no no
I'm asking what the relationship between $vol(B^n)$ and $vol(S^{n-1})$ is
I've forgotten
Huy
Huy
ah
can't you just generalize the integral
dV = A * dr basically
uhh
user227867
Basically is an overused word.
14:56
these are unit spheres and balls btw
Huy
Huy
that'd be the physicist proof
so dV/dr = A
...
Huy
Huy
wat
that's the relationship
that's not an answer
Huy
Huy
???
14:57
ugh I once wrote down this proof
gotta find it now
Huy
Huy
I just wrote it down
???
???
what does that mean
Huy
Huy
that you're confusing me
ah, here it is
I have a full derivation of the induced metric on a sphere
computation of the curvatures in all dimensions
Riemann, Ricci, scalar
pretty cool
uhh
what kind of crappy notation is this
Huy
Huy
15:05
lol
you didn't say you wanted that formula
you said the relationship between volume of B and S
...
yeah
hmm
what do
I really have no clue what these symbols mean
have to decode them
Huy
Huy
wat
which one
ö_ö
I think $\mathcal V(n,r)$ is the volume of the $(n+1)$-ball
and $\mathcal A(n,r)$ is the volume of the $n$-sphere
@JasperLoy JASPER!!! I wanna listen to more songs from you!!!
So what I need for this proof is the ratio of the volume of $S^{n-1}$ to $B^n$.
which is...
n?
15:08
@JasperLoy I saw the message you let in Tex chat, please don't call me genius anymore. If I were a genius I would have probably taught mathematics at MIT, Princeton, Harvard from an early age as a professor.
Huy
Huy
why n
wait, no
hmm
yes, that's right
u taught in harvard ?
compute $\mathcal A(n-1,1)/\mathcal V(n-1,1)$.
Huy
Huy
no
15:09
it's $n$.
Huy
Huy
ok
@Agawa001 No. But a genius does it.
many geniuses exist in this world as grigory (he refused a nobel price and never taught in a world wide recongnised institute)
@Huy I'm trying to show that $$\operatorname{tr}_gT=\frac{1}{\operatorname{vol}S^{n-1}}\int_{S^{n-1}}T(V,V)\,‌​d\sigma(V)$$
Grigory Perelman
Huy
Huy
15:12
ok?
that factor of vol S is killing me
What is $T(V, V)$
tensor $T$ acting on $V$ twice
@Agawa001 I think so. Thinking freely without the bounds imposed by any insitution is one of the most powerful way to produce mathematics.
Without being pushed to publish because it is required so.
The inner manifestation of a free spirit in mathematics is what you want if you wanna obtain the results Ramanujan or other important figures of mathematics obtained. You have no bounds.
manifesto
keyword
user227867
15:17
@user1618033 I now have 4 videos on my channel. I made 2 today.
@JasperLoy let me see!
user227867
@user1618033 OK. But those professors are not genius. You are. =)
Oh that's only for a symmetric tensor
user227867
@user1618033 Click link, as usual. =) Don't post in this room, please.
@JasperLoy would you bet those professors aren't geniuses?
user227867
15:19
@user1618033 Yes, I bet they are not as genius as you. =)
@JasperLoy lol
@JasperLoy I'm listening to sole mio.
Great song!!!
user227867
@user1618033 Some people may think I am sucking up to you by praising you. Not true. I truly admire your mathematical talent.
@JasperLoy Thank you. But don't overly appreciate me. ;)
user227867
Ethan is also another genius, but he doesn't come to chat these days. He is busy with school.
(I still have a lot of stuff to study)
I didn't see him either.
user227867
15:21
On the other hand, I am not a genius. So I need to work very hard just to write an article in future.
@JasperLoy You really have a great voice! Mayb you should do more than only having these posts on youtube.
user227867
Sometimes, I think, if I love mathematics so much, maybe I should not be a professor, maybe I should just read my books all day long.
@JasperLoy don't forget that I've always said I worked extremely hard for all I got.
user227867
But I think one needs a job, a career, so being a professor is not a bad thing.
@JasperLoy That's true.
15:23
hello, someone knows how to write this with latex
user227867
@user1618033 A mad person emailed me to scold me this morning.
@JasperLoy Why? What was it about?
user227867
@user1618033 This person probably thought I was ignoring him in chat, when all I did was go to sleep. Never mind, I will ignore him from now. He abused me years ago and did not apologise, and today he abused me again.
Huy
Huy
$\overline{L^\infty(\Omega)}^{\| . \|_{(\Phi)}}$
@JasperLoy Sorry to hear that. From this chat??
user227867
15:26
@user1618033 Yes, it is someone on Math SE.
@JasperLoy pfff. That's awful.
@Huy thank you
good lord what is that
@JasperLoy Some dates lately?
:D
user227867
@user1618033 Sorry, what do you mean? I have no dates, no girlfriends. =)
15:36
@JasperLoy That's strange to me. If you go out in the street and sing sole mio as in the youtube clip, I can predit that a dozen of girls fall down at your knees in the shortest period of time.
:-)
user227867
@user1618033 LOL. If you go out on the street and solve your integral, the same will happen. =P
@JasperLoy hahahahaha :-))))
@JasperLoy NEVER THIS WAY. JUST TO SCARE THEM! ;)
user227867
@user1618033 When I was in high school, I had a genius schoolmate. For a few hard integrals, he could come up with several antiderivatives for each of them that nobody could think of. I copied them down and went home to study the solutions.
@JasperLoy do you have in mind such an example to see it now?
user227867
@user1618033 I have forgotten what they are. =) But it is not the same level as what you do. I think you can solve those when you were 5 years old. =)
15:43
@JasperLoy When I was 5 years old I think I was doing something else, playing some game? :D
user227867
@user1618033 Like having a date. =)
@JasperLoy :D
user227867
Have you heard of the Chinese mathematician Hua Loo Keng or Hua Luo Geng?
user227867
I think in China, there is a school named after him, and they train their students in mathematical olympiad problems from grade 1 on wards.
@JasperLoy It doesn't sound familiar to me although I might have heard about it before.
Hua Luogeng, or Hua Loo-Keng (Chinese: 华罗庚; Wade–Giles: Hua Lo-keng; 12 November 1910 – 12 June 1985), was a Chinese mathematician famous for his important contributions to number theory and for his role as the leader of mathematics research and education in the People's Republic of China. He was largely responsible for identifying and nurturing the renowned mathematician Chen Jingrun who proved Chen's theorem, the best known result on the Goldbach conjecture. In addition, Hua's later work on mathematical optimization and operations research made an enormous impact on China's economy. Hua did not...
user227867
15:46
I don't have their textbooks with me now, but they were written in Chinese and are very very good. 12 books, one for each level.
@JasperLoy Are there English versions too?
user227867
@user1618033 Yeah, he has this beautiful Springer book 'Introduction to number theory'
@JasperLoy hmmm, thanks! Let me check that.
user227867
@user1618033 Unfortunately, no. I can't find a complete set of those books now either.
@JasperLoy OK. Let me know if you ever find English versions, I'm interested.
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