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12:48 AM
One of the exercises in my physics course discusses the "maxwell-jeffrey model". Googling these did not bring up the results I intended to find
 
lmao
 
 
1 hour later…
1:57 AM
I propose deprecating the phrase "rectangular prism". It should be replaced with "geometric box", or simply "box" if context makes clear that one is referring to the shape rather than the real-world objects
(I reject "cuboid" as it is too ambiguous)
 
yeh thats a box
I support your cause
 
it's a ball
actually, what about polyinterval
 
typical algebraists
 
bi-triangular prism, or if greater precision is needed, bi-right-triangular prism
 
I swear the terminology of algebraic topology, geometry and number theory needs to be decluttered as a whole
set theorists "we have these incredibly complicated core models... let's call 'em 'mice'"
 
2:13 AM
cubical simplex
 
algebraic geometers: "hmmm... I think crystalline cohomology is a fine term"
 
"Remark 14.61. The above proof does not provide an explicit bound on q(L). Boris Weisfeiler [Wei84] proved that for n > 63, q(GL(n;C)) 6 (n + 2)!, which is nearly optimal since GL(n;C) contains the permutation group Sn which has the order n!. Weisfeiler obtained this result in 1984, shortly before he, tragically, disappeared in Chile in 1985. On August 21 of 2012 a Chilean judge ordered the arrest of eight retired police and military officers in connection with the kidnapping and disappearance of Weisfeiler.
that remark got out of hand quite fast
 
US-funded death squad reaccs only
 
holy cow
 
also I really wonder why copypasting the lesser or equal than symbol from the pdf turned into a 6 here
works in mysterious ways
 
2:18 AM
F.U / !
F.U i / ⇒
F.U i \ U j /;
To facilitate searching. Clearly.
A comma was replaced with a semicolon at the end.
25
A: In which way have fake spaces made it to actual use?

Steven B. SegletesIn reference to your point, In particular, has copy-paste of mathematics been implemented so that it may be used without disadvantages I don't know the particulars of what you cite, but the accsupp package allows different things to be displayed in a PDF versus what shows up in a copy paste. R...

 
If I have a rational laplace transform of the form $\hat{Z}(z) = \frac{izm-1}{\tau m(z+i(y-x))(z + i(x+y))}$, with $-(x+y)i, -(y-x)i$ poles on the negative imaginary axis, is there a smarter way to calculate the inverse laplace transform than doing a contour in an ever-larger half-circle
I could possibly nuke this and do a 3-times convolution of two exponential functions with a dirac delta + its distributional derivative but I think that's not the intended solution
And somehow just adding up the residues gives stupid results. I should probably just check that calculation but I don't see where I went wrong
*2-times convolution
 
2:56 AM
what's stopping you from just doing partial fractions
It's something of the form $\dfrac{A}{z-z_1}+\dfrac{B}{z-z_2}$
 
3:42 AM
Hm yeah that does yield sensible results
We were told to do it via residues but screw that
I'm happy as long as my solutions are real-valued
 
well, i wouldn't expect it to be in this case. Z(z) isn't real on the real line
 
Our laplace transform is defined as the integral of Z(t) * exp(izt) dt for t in [0,inf) and z with positive imaginary part
 
ahhh
so more like a unilateral fourier transform (in my book)
 
@MikeMiller Yes, I think so. I don't know any applications though.
 
This is superficially different than other definitions and I think this is the reason why my other calculation doesn't add up
 
3:49 AM
yeah, probably
though, do note that partial fractions facilitates residues
 
Because I literally guessed how the inverse laplace formula changes in this formula
And with the partial fractions I can circumvent the inverse laplace because i know that 1/(iz + a) is the laplace transform of e^at
(with our def)
 
nice
i mean, $\text{res}(\frac{f(z)}{z-a},a)=f(a)$ if $f(z)$ analytic at $z=a$ is simple enough
 
it is, the residues are super easy to calculate, which is why I was mad
 
ah, lol
 
And the rest of the proof, like proving that the integral over the arc in the contour integral goes to zero, is also not too hard. But I'll just ask the prof for the actual formula of the inverse laplace and I'm sure then it'll work out just fine
 
4:20 AM
I am planning to read curve complexes next semester, and looking for some motivating source, any help?
 
I don't know a source but the point is it's a negatively curved complex on which MCG acts geometrically
There are some short proofs of hyperbolicity of the curve complex which might be interesting to read, look up Alessandro Sisto's blog
 
4:54 AM
@EdwardEvans @Astyx @Alessandro: youtube.com/watch?v=HO55ibNgN1U
 
underrated masterpiece
listens to it at 1.75 speed
 
lool
 
5:38 AM
Hi @Ted!
 
Hi, a Balarka.
 
You were right about the centers! That's pretty surprising.
 
Not really. $If b/a=c/b$, with $a<b<c$, then $c-b>b-a$ always :)
 
I never remember formulas, so I had to compute.
 
I've also done this stuff in the last lecture of my diff geo.
Well, lengths on vertical rays are easy. Distance from $(0,a)$ to $(0,b)$ is $|\log b/a|$. That's not even a computation.
 
5:41 AM
Yeah.
In the Poincare disk model the center is shifted towards the ideal boundary, so I thought it'd be shifted towards $\infty$ sometimes (the only point in the ideal boundary not present in the upper half plane model), hence the confusion.
 
That's not a point worth considering. :)
 
But somehow all the "mass" in the ideal boundary is on the x-axis, so $\infty$ doesn't matter enough to pull the center towards it. Yeah.
 
I don't do enough with the disk model (other than complex analysis stuff); I have taught the upper half plane so many times in diff geo it's now ingrained in me.
And, no, I'm not teaching the undergrads about the quadric in Lorentz space.
 
Haha
 
 
2 hours later…
8:07 AM
HI.
Sorry, I have some stupid questions regarding mathematical education.
So, you see, there are multiple ways to approach mathematics. One could simply read everything in a book and follow along, or one could try to derive the formulae themselves, or even explore why exactly a particular thing in mathematics is the way it is, why a topic came into existence, what it's applications/implications are etc.
But how should one really study mathematics? What would be the "perfect" way to do so? Is it something subjective or objective? How exactly is someone who derives the formulae themselves and
3
 
It's easy to deceive oneself into thinking they understand something, especially in the beginning because one needs to spend some time thinking, reading, and above all listening other people talk about mathematics to know that there is such a thing as a "standard of quality" for understanding in the community. Understanding is not inherently an objective word: from personal experience I have felt I understood a particular thing many times, and then a few weeks later realize I still don't get it.
 
Welcome @Norman_22194
 
@BalarkaSen In that case, how exactly do you determine whether you've understood something or not?
@user147593 Thanks! :)
 
There is no perfect way to study mathematics but there are a few "community guidelines" to ensure one understands a particular thing, as per the standards of the community. This includes reading and understanding proofs, doing exercises pertaining to the thing, etc.
In the beginning most of the things you'd want to understand is digested very well by centuries that led up to modern mathematics, so there are well-written textbooks with properly exposed proofs written, and ample exercises provided, so that you can do both.
 
@BalarkaSen I don't get it. I don't see how doing exercises properly implies that you have a good understanding...
 
8:20 AM
your question amounts to what is known as "mathematical maturity"
 
@Norman_22194 It's part of the definition of what understanding means, in the community. To be able to do exercises implies you can use the ideas in the textbooks, which is considered to be a necessary condition for understanding the ideas.
 
@user147593 Wikipedia : "It pertains to a mixture of mathematical experience and insight that cannot be directly taught." :-/
3
 
There is nothing to get, it's how "understanding" is defined. It's not an objective word, so people have defined it to be a certain thing.
 
what do you think the difference is between "learning" and "understanding"? @Norman_22194
 
memory vs model.
 
8:24 AM
@user147593 I'm too confused to differentiate between the two. I can't even say that I understand the meaning of the two terms...
@user147593 To me, understanding means to be able to grasp the idea of something.
And learning means to acquire skill via being taught, or experience.
 
one is an action, one is a state?
 
@copper.hat well, understanding could be treated as a verb, no?
 
please think about it carefully
How does one depend on the other?
 
"The idea of something" is meaningless because who knows if what you think is the idea behind something is actually the idea behind it?
 
@BalarkaSen It's subjective so "actually the idea behind it" shouldn't exist, should it?
 
8:27 AM
One way mathematicians gauge if some idea (which is really a collection of tricks or even better, techniques) is more important over another is if they can be applied to prove new things
 
i think you need to be more specific to have more than a fun conversation :-)
 
6 mins ago, by user 147593
what do you think the difference is between "learning" and "understanding"? @Norman_22194
 
@copper.hat Um, can you please tell me where I need to be more specific? I would really like to have a better conversation.
 
3 mins ago, by user 147593
How does one depend on the other?
 
@user147593 I was still typing a reply to that :)
 
8:30 AM
:)
 
Alright, so here's what understanding (a mathematical concept) means to me :
It means to be able to grasp it's idea, to understand why it's required, how it can be used to solve problems, what led to it's origin, where it can be used, how this helps us in understanding things better etc
And learning, to me, is to acquire knowledge about something by means of being taught or from experience gained by dealing with a particular topic for a long amount of time.
(I know that this will contradict what most people understand both of these terms as 😅)
 
so would i, but bedtime calls, have some fires to fight in the morning..
 
@copper.hat Good night! :)
 
gn! enjoy!
 
It's vs its @Norman_22194
 
8:35 AM
@Norman_22194 Well for example you can say the idea behind polar coordinates is to write everything in terms of trigonometric functions. This is a useless idea because it doesn't help you to do anything with it. A better thing to say is that idea behind polar coordinates is to exploit radial symmetry. But you can do better, maybe
There's no philosophical solution to perfect understanding.
 
Yet, perfect learning is to learn by heart.
 
@BalarkaSen In such situations, I agree that "an actual idea" does exist...
 
If you can use your "idea" to prove many things and proofs of known things become simpler it's pretty good. This is the general rule of thumb.
 
@NoName "I’m playing both sides, so that I always come out on top."
@BalarkaSen This, I agree on...
Alright, so here're the conclusions I've arrived at, so far :
There indeed are multiple ways to approach mathematics that fulfil their own purposes. In the mathematical community, the approach that could be defined as good should be such that it gives rise to ideas that can be used to prove many things and make proofs of some known things simpler (and maybe even involve some intuition).
And a perfect method $\not\exists$.
So, did I misunderstand something?
 
The first sentence captures it best.
 
8:45 AM
LOL
 
:D
@BalarkaSen By the way, did you go through my questions?
 
i had a look to understand what could be a good example
 
That's nice.
Thanks, everyone :)
 
@BalarkaSen I was in consultancy for two years after a maths degree, now I'm doing a PhD in crypto.
 
cya
 
8:49 AM
ahh ok @Krijn
 
Take care :)
 
That was really valuable, because you learn so much from skills you didn't know you have, and also because you know what you really want to do and what not @EdwardEvans
I also spoke about this in a podcast of some friends a few weeks ago if that helps anyone
(although I do have to relearn things I already learned in my masters, such as parts of algebraic number theory, algebraic geometry and commutative algebra. So that's definitely a downside)
 
@Norman_22194 Learning≠Memorising?
 
Rote learning does.
 
@Thorgott Yes, but currently, with isogeny-based cryptography it is getting into some really really cool mathematical stuff as well while at the same time applying that directly in real-world applications
 
8:54 AM
@YouKnowMe Maybe. I don't think just memorization involves any understanding whereas learning does.
 
Rote learning is a memorization technique based on repetition. The idea is that one will be able to quickly recall the meaning of the material the more one repeats it. Some of the alternatives to rote learning include meaningful learning, associative learning, and active learning. == Versus critical thinking == Rote methods are routinely used when fast memorization is required, such as learning one's lines in a play or memorizing a telephone number. Rote learning is widely used in the mastery of foundational knowledge. Examples of school topics where rote learning is frequently used include phonics...
 
"More than ever, mathematics must include the mastery of concepts instead of mere memorization and the following of procedures. More than ever, school mathematics must include an understanding of how to use technology to arrive meaningfully at solutions to problems instead of endless attention to increasingly outdated computational tedium."
"Newer standards often recommend that students derive formulas themselves to achieve the best understanding."
Interesting stuff.
 
yup, there are some good references at the bottom also
 
Thanks for sharing :)
 
9:02 AM
@user147593 $\neq p$ XD
 
Rote learning is useful for things like medicine.
 
@Norman_22194 prove it
 
@NoName Right, but I still think it still involves some level of understanding.
@KarlKroningfeld Let $n=2$ and $p=1$. So, $np=2\neq1\implies p\neq np$. Easy peasy.
 
Yeah, of course. You've to know what you're memorising. But I was shocked by the amount active memorisation involved in medicine degrees. Once I saw the sheer amount of information they are expected to retain, it all made sense.
 
@NoName True, my father (a former veterinarian) still remembers some info related to medicines that he learned about decades earlier.
 
9:07 AM
@Norman_22194 Nice.
 
@KarlKroningfeld Don't worry, you can have the million dollars.
I need to go now, cya later.
 
 
4 hours later…
1:06 PM
Should I feel bad about asking questions? I feel like my questions are annoying people, as they're basic and wrong. :(
 
1:18 PM
you're always free to ask
as long as you don't spam the same question and randomly ping people
If people are interested in answering your question, they will
 
I meant on the site, not in this chat.
 
The same applies there. Just make sure to check if the question hasn't been asked yet, to include context and that the question is clear and shows you've done some work on your own
 
I try to do that.
 
You shouldn't feel bad about asking basic question. Everyone starts somewhere. However if you ask a question because you were too lazy to do the work yourself, then it wastes both your and the answerer's time, since you probably won't learn as efficiently
 
Well, all of my questions are about confusions about theory, rather than problems that I am stuck on. I try to list examples of my confusions as well.
 
1:30 PM
Q46
Please help
A step by step guidance for such questions is what I am looking for
 
Have you managed anything so far ?
 
@Astyx Me?
 
yes
 
All that I know is this formula
I do not know what the final term will be here?
 
hullo
 
1:40 PM
hi
which of these terms are you given in this problem ?
 
n(A U B U C) and intersection is not known in terms of probablity
Else everything is known..
Am I correct...
???
 
Can you write explicitly which is which ?
 
1:55 PM
I feel n(A)=n(B)=n(C)=0.75;Intersections equal 0.50
 
not quite
If I have a 75% chance of passing maths and 75% chance of passing physics, then I have a greater chance than 75% of passing at least one of them
 
@Astyx Could not get you
Can u elaborate slightly
ok
@Astyx I got the point
What next
 
So you need to see what that 75% is
 
Exactly
What is that?
 
In which case do you not pass at least one ?
 
2:06 PM
@Astyx I do not know..
 
If you fail them all right ?
 
ok so far
 
Tim
Good day! Does anyone have access to this book? doi.org/10.1145/3335772
 
Ok, now with the same thinking, can you link the probability of passing them all with the data that is given ?
 
No trying hard..
But not ablle to
 
2:13 PM
Ok let's try another way
Let $p_3$ be the probability of passing all 3, $p_2$ of passing 2, etc for $p_1$ and $p_0$
 
2:31 PM
ok then...
'?
 
Can you link those to the given data ?
 
No not really
I am having problem with exactly2 and atleast2,exactly1 and atleast 1
I do not know how to relate these things in the venn diagram
@Astyx
 
If you pass 3, do you pass at least 2 ? If you pass 2 do you pass at least 2 ? etc ?
 
Yes
btw that should be atleast one
 
2:48 PM
So write the equations down
 
3:00 PM
I am getting p+m+c-(0.4X3)=p(m U p U c)-p(m intersection p intersection c)
I do not know how to comprehend RHS
@Astyx
 
No, I want you to link them to p_1, p_2 etc
 
@Astyx I do not know how to comprehend 40% as?
Please explain this to me
 
If that seemed too much like cheating, you could also prove it directly using induction. We claim is that if $A$ is a linear transformation on $V$ that is nilpotent of order $m$, then $\operatorname{dim} V \geq m .$ The base case $m=0$ is trivial.
So, assume the claim is true for $m-1,$ and let
$$
W=\left\{x \in V \mid A^{m-1} x=0\right\}
$$
a.k.a. the null space of $A^{m-1}$. It is easy to check that $A$ maps $W$ into itself, and $A$ is nilpotent of order $m-1$ on $W$. By induction, $\operatorname{dim} W \geq m-1$. But since $W$ is a proper subspace of $V,$ it follows that $\operatorname{d
 
40% is exactly p_2, by definition of p_2
 
Is this proof correct for Nilpotent operator/matrix: A operator $T$ on a vector space $V$ is called nilpotent if $T^{k}=0$ for some positive integer $k$. The positive integer $m$ such that $T^{m}=0$ and $T^{m-1} \neq 0$ is called the nilpotency index of $T$.
(a) Prove that if $m$ is the nilpotency index of $T$, then $m$ cannot exceed the dimension of $V$.
 
3:06 PM
Then how to deal with atleast
I do not understand
@Astyx
 
If you pass at least 2, it means you passed exactly 3 or you passed exactly 2
 
Is this proof correct for Nilpotent operator/matrix: A operator $T$ on a vector space $V$ is called nilpotent if $T^{k}=0$ for some positive integer $k$. The positive integer $m$ such that $T^{m}=0$ and $T^{m-1} \neq 0$ is called the nilpotency index of $T$.
(a) Prove that if $m$ is the nilpotency index of $T$, then $m$ cannot exceed the dimension of $V$.
If that seemed too much like cheating, you could also prove it directly using induction. We claim is that if $A$ is a linear transformation on $V$ that is nilpotent of order $m$, then $\operatorname{dim} V \geq m .$ The base case $m=0$ is trivial.
So, assume the claim is true for $m-1,$ and let
$$
W=\left\{x \in V \mid A^{m-1} x=0\right\}
$$
a.k.a. the null space of $A^{m-1}$. It is easy to check that $A$ maps $W$ into itself, and $A$ is nilpotent of order $m-1$ on $W$. By induction, $\operatorname{dim} W \geq m-1$. But since $W$ is a proper subspace of $V,$ it follows that $\operatorname{d
Is this proof mathematically correct ? For part (a)
 
@Astyx Or at least 1
But there could be a situation where the guy could not pass even one..
 
Yes, that's p_0
@mathsstudent Looks fine to me
 
Atleast 2-exactly 2 is what
Exactly 1?
But I do not think so
@Astyx
 
3:17 PM
No, think about it. If I give you at least two (out of three) balls, but not exactly two, how many balls do I give you
 
I feel it should be P_0 or P_1..
I mean should not be exactly 1 @Astyx
 
i don't understand what you mean
 
Can u solve this problem for me..
I have got the intution
 
So say we have $A$ a domain, integrally closed in its field of fractions $K$, $L/K$ a separable extension of degree $n$, and $B$ the integral closure of $A$ in $L$. Now fix and algebraic closure $\overline{K}$ of $K$. Why are there $n$ distinct embeddings of $L$ into $\overline{K}$?
 
@Astyx But without the steps I will not understand..
 
3:24 PM
@BigSocks that's just what it means to be separable
 
yeah, rereading what I wrote it's kind of obvious
y e e s h
 
Consider the statement: "Prove or disprove: Every even integer is the sum of three distinct even integers."

Why is it incorrect to say that "This statement says that the sum of three distinct even integers is even"? That is, can we begin by letting n = a + b + c for even integers n, a, b, c and then letting a = 2x, b = 2y, c = 2z for integers x,y,z?
 
$n$ distinct embeddings bc you have $n$ distinct conjugates of an element in the closure and each conjugate is a different embedding
just kept thinking of separability in terms of derivatives being nonzero. gotta keep the other notion in mind
 
@Astyx I got two separate values
I do not know how to solve it
Like 0.5-0.4 gives me 0.1
Now that should supposedly be the value for atleast 1
Isnt it
Howver the value in my question reads 0.75
Why are they not matching??
Please explain@Astyx
 
@BigSocks I think you can say that every element of the Galois group gives you a different embedding, right?
 
3:38 PM
yeah that is what I originally thought would be the answer, but I wanted something closer to definitions
I was also wondering about whether or not there would be "extra" extensions you could get bc the algebraic closure could be quite large, but it helps to remember that these $n$ embeddings are really talking about $L$ as an extension of $K$
 
@politeproofs Because you then only show that the sum of $a$, $b$, and $c$ is even, not necessarily that an even number (say 2) can be written as such a sum
@BigSocks Yeah you need to fix $K$ elements so that doesn't leave too much room
Would be how I think about it intuitively
 
How do I get the latex to work in chat
I clicked on the link and ran it
 
you put it in bookmarks?
 
And then press it
 
with the chat open
 
3:46 PM
Cheers <3
 
@PrinceNorthLæraðr If u wish u can reply me here
 
I'm not really a person who gives math help as much as receives it :)
 
The question is already posted
 
I'm not sure I could help you, sorry
I also have class starting in 7 minutes so
Sorry about that
 
4:08 PM
I am so happy that censorship has helped me delete my illegal personality traits
 
I need to find PDF of X where f(x,y)=y^2*e^(-y(x+1)) but the integral is too complicated
What else can i do?
Where x>0, y>0
 
@Krijn Thanks for the info :)
@BalarkaSen nise
 
4:25 PM
@Balarka this is good!
 
yeah
 
@Eminem you mean the integral from 0 to infinity of f(x,y)dy?
Have you tried partial integration?
 
@Krijn there was no normality assumption
 
Ah, yeah, sharp. Missed that
 
@user2103480 I have. Does not work...
 
4:34 PM
@Eminem try again
 
Why does the following proof work for this question: If $a,b,c$ are any three distinct positive integers such that $1/a + 1/b + 1/c = 1$ then $a + b + c$ is prime.

Let $a,b,c$ be three distinct positive integers such that $1/a + 1/b + 1/c = 1$. WLOG $a < b < c$ so $1/a > 1/b > 1/c$. So $1/a > 1/3$ so $a = 2$. Since $1/b > 1/4$, and $b \neq 2$ means $b = 3$. So $1/c = 1 - 1/2 - 1/3 = 1/6$ so $c = 6$. So $a + b + c = 11$ which is prime.
But isn't that just a proof by example, instead of proving it for all positive distinct integers?
 
proof by there are no other examples
 
@LeakyNun Hm, I don't follow. Just because $1/a > 1/3$ does not mean $a = 2$ necessarily or?
 
proof by confusing students
 
what else can a be
 
4:40 PM
@LeakyNun Why not $3$? Why not $5$? Why not $20$? Why instead $2$?
 
1/a > 1/3
so a < 3
 
Oops.
Oh well, that's my silly mistake of the day I guess.. Thanks.
 
5:08 PM
@Balarka very cool album, Ne Obliviscaris vibes
 
5:18 PM
Hello!! Could someone of you have a look at my question about convergence of integrals?
0
Q: Convergence or divergence of improper integrals

Mary StarI want to check if the following integrals converge or diverge. $\displaystyle{\int_1^{+\infty}\frac{(\ln t)^{\beta}}{t^{\alpha}}\, dt, \alpha ,\beta\in \mathbb{R}}$ $\displaystyle{\int_0^{+\infty}\frac{t\ln t}{(1-t^2)^2}\, dt}$ $\displaystyle{\int_1^{+\infty}\frac{|\sin t|}{t}\, dt}$ $\disp...

 
Given any two groups G,H - the direct product is necessarily not simple?
I was reading math.stackexchange.com/a/3195900/778005, and it was kind of implied from the answer
 
(G x H) / (G x {1}) = H
 
if both factors are non-trivial, of course
 
:|, yeah sorry, it's obvious, don't know why I was confused
 
5:42 PM
Say we have a setup with $A$ an integrally closed domain over its fraction field $K$, $L/K$ a separable extension of degree $n$, $B$ the integral closure of $A$ in $L$, $\{e_1, ..., e_n \}$ a basis for $L/K$ contained in $B$, and $\overline{K}$ a fixed algebraic closure of $K$. Also we have $n$ distinct embeddings of $L$ into $\overline{K}$, just like we said above. We call them $\sigma_i$

For $\alpha \in L$, we write down $\alpha$ in terms of the basis elements times some elements of $K$, namely $x_1 e_1 + ... + x_n e_n$. Now some questions
At $2.$ it should say $\alpha \in B \Rightarrow \sigma_i (\alpha)$; forgot to capitalize
Similarly for the $\rightarrow$ at $3.$
fwiw I am looking at Lorenzini- An invitation to Arithmetic Geometry. Also $d$ is $(det(M))^2$ of $M$ the matrix of all the $\sigma_i (e_j)$
 
@Thorgott Similar question - if G,H are non-solvable groups, then their direct product is also non-solvable? this, I hope, less obvious :D
 
what do you know about subgroups/quotients of solvable groups?
 
@Thorgott Ah, you're saying that G x {e} is solvable and isomorhpic to G?
 
@BigSocks 1. is nonsense, you probably meant to write something different
@yong yes
 
Damn, I need to sleep
@Thorgott Thanks!
 
6:05 PM
Right, I meant to say “since it’s the integral closure of $A$ in $L$”
 
no, I mean the claim is nonsense
 
$\alpha \in B$ you mean?
 
yes
why would an arbitrary element of L lie in B
 
weird, the book says "Since $\alpha \in B$, each $\sigma_i (\alpha)$ is integral over $A$
yeah, that's what I was wondering too, man
maybe they mean to put "Since $\alpha \in L$" idk
but then idk why the implication would hold
maybe it's an error I guess? but it seems like a big error bc it helps most of the argument
 
perhaps there was an additional assumption such as the $x_i$ being in $A$?
 
6:11 PM
hmm no, they are for sure just in $K$
thanks for thinking about it though. it has got me stumped
the basis can be turned into a basis for $L$ with some choice of elements in $A$ though so
that was a corollary some time ago
 
I thought you already have a basis for $L$? that's what you said
 
yeah yes
but it's just elements of $B$
idk, just skimming through the book to see if there's something I missed. They do assume the basis is there, it's not that I constructed it in some way or anything
(but yeah I was just saying that you always can construct it I guess)
(and you get $L$ the field of fractions of $B$)
 
yeah, $B$ generates $L$ as $K$-vector space
in any case, something is wrong there, but I don't know what
 
ok well at least it makes it more interesting- now trying to also fix the argument as I understand it
doesn't seem to be here
it's proposition 4.8, pages 21-22
 
 
1 hour later…
7:24 PM
0
Q: What is going on in this proof about finitely-generated modules?

BigSocksSo this is proposition $4.8$ in Lorenzini's "An invitation to arithmetic geometry": Let $A$ be a domain, integrally closed, in its field of fractions $K$. Let $L/K$ be a separable extension of degree $n$. Let $\{ e_1, ..., e_n \} \subset B$ be a basis for $L$ over $K$ where $B$ is the integral cl...

I posted it here
 
oh, you omitted an important piece of information
 
oh yeah?
 
he considers $\alpha$ integral over $A$, of course it is in $B$ then
 
wot
 
""We need to show the existence of a nonzero element d such that dxi∈A for all i=1,...,n, whenever α=Σni=1xiei is integral over A.""
 
7:28 PM
I mean I guess that was my question 3
yeah I guess $\alpha$ integral over $A$ implies $\alpha \in [$ integral closure of $A ]$
 
Hey @Thorgott
 
to answer $Q1$, if $\alpha=\sum_{i=1}^nx_ie_i$ and $dx_i\in A$ for all $i$, then $\alpha=\sum_{i=1}^ndx_i\frac{e_i}{d}$ is an $A$-linear combination of the elements, $e_1/d,...,e_n/d$, so contained in the free $A$-module generated by $e_1/d,...,e_n/d$
hey @Alessandro
 
ah ok kinda easy
 
Sanity check: I have an inverse system $X_i,f_{ij}$ of topological spaces with limit $X$, such that every $f_{ij}$ is a retraction. Is it true that the projections $\pi_i\colon X\to X_i$ are retractions? I think so, because if $i_{ji}:X_j\to X_i$ witnesses that $f_{ij}:X_i\to X_j$ is a retraction (meaning that $i_{ji}\circ f_{ij}=\mathrm{Id}_{X_i}$), then the map from $X_i$ to $X$ that is "the sequence of the $i_{ij}$" should witness that $\pi_i$ is a retraction
 
for $Q2$, three general facts:
- homomorphic images of integral elements are integral
- sums of integral elements are integral
- products of integral elements are integral
 
7:41 PM
mmm- homomorphic images of integral elements are integral... how do I see this? I guess it's related to "being integral" caring more about the coefficients being in $A$ rather than the rest of the term... maybe
 
homomorphisms commute with polynomials
@Alessandro I'm skeptical. You'd have to somehow convince me that you can choose sections in a manner compatible with all the retractions simultaneously and this isn't clear to me.
 
inverse system is a sequence for me
I'm not dealing with arbitrary limits
 
hmm ok, then it sounds more plausible
 
I guess what I want to say is that really $X=\bigcup X_i$ in the way one would expect
 
then all we have to do is look at the maps $X_{i+1}\rightarrow X_i$. Choose a section for each. Fix an index $i$. For each $j$, we have a map $X_i\rightarrow X_j$ that's the composition of the retractions if $i\ge j$ and of the sections if $i\le j$. Then this is compatible with all the retractions $X_{j+1}\rightarrow X_j$ by construction, so induces a map $X_i\rightarrow X$ by universal property, which is a section of the projection $X\rightarrow X_i$ by construction.
pure algebra
 
7:54 PM
nice
Now I'm curious whether this works for all limits of topological spaces but it seems ugly in general
 
the answer is no for very trivial reasons
product with empty set
it's definitely an interesting question for filtered limits
it also fails for less trivial reasons in general, the projections from fiber products are in general not surjective, hence not retractions
 
So @Thorgott is it fair to say that if you have $f(\beta)$ witnessing the integrality of $\beta$ in the ring that contains the coefficients of $f$, then $\sigma (f(\beta) = f (\sigma (\beta))$ witnesses the integrality of $\sigma(\beta)$ in the same ring of coefficients?
 
precisely
 
neat ok - getting through this
thanks again, of course
 
8:10 PM
np
@Alessandro can you think of a commuting square of retractions such that you can't find a commuting square of corresponding sections?
wait of course you can't, I'm silly
ok, obviously if the directed set contains a cofinal sequence, my above argument applies, so that can be excluded
 
@EdwardEvans I just checked your soundcloud, no new releases?
 
The notion of a rational equivalence on cycles seems very similar to unoriented bordism of two manifolds. Is there a categorical similarity between the two?
 
@Semiclassical I mentioned to the prof that one could do the inverse laplace via partial fractions and he was like "ah yeah that's probably easier, I always forget how that works" lol
 
8:38 PM
Is there a Marhematica package or otherwise for calculating the Fourier coefficients for a theta form associated to a quadratic form?
 
8:57 PM
I do see that some folks have defined algebraic cobordisms. I don't know how much rational equivalence comes there.
 
EM4
Question

$\Bbb{Z} \cap [0,6)$ is {0,1,2,3,4,5} that right.
 
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