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12:05 AM
well, $\Bbb F_{p^d}$ in the notation of that question
 
Good point.
@LukasHeger so then it follows that the minimal polynomial of a root over $\mathbb F_{p^d}$ has degree
$$\deg m_{\alpha_i,\mathbb F_{p^d}} = [\mathbb F_{p^d}(\alpha):\mathbb F_{p^d}] = [\mathbb F_{p^{lcm(d,m)}}:\mathbb F_{p^d}] = \frac{lcm(d,m)}{d} = \frac{m}{\gcd(d,m)}?$$
Sorry for the pinging for edits.
How did you end up getting gcd(d,m) as the degree?
(or rather, where did I go wrong)
 
you're right, each factor should have degree $\frac{d}{\gcd(d,m)}$
you want $\Bbb F_{p^m}$ there not $\Bbb F_{p^d}$ though
so much for my experience :P
at least the proof method was correct
just didn't do the calculations correctly lol
 
Why $d$ in the numerator and not $m$?
 
because you switched up $d$ and $m$
we want to look at the minimal polynomial over $\Bbb F_{p^m}$
 
OH
Okay, makes sense.
And then does the number of polynomials which $f$ breaks into change as a result?
 
12:15 AM
yeah, $\mathrm{gcd}(d,m)$ of those
so I just switched the number and the degrees of the polynomials in the answer
 
And that's because we know they are all of the same degree and the degrees will add to $d$.
 
right
 
Great! Thanks!
btw, is there a nicer notation for minimal polynomials which is not as clunky looking?
 
I'm not aware of any
 
Hmmm, alright.
Now to think of some tricky partial derivatives to mess students up.
Can derivatives ever be truly tricky.
 
12:33 AM
they can be ugly
 
That's very true.
 
12:44 AM
Can I tell an awful joke
 
1:07 AM
If it is just a "bad" joke, then sure.
If it is supposed to be offensive or something like that, then probably not.
 
1:21 AM
@anakhro Why does Marry Poppins have painful period cramps
Supercalifragilistic-endometriosis
 
It was my understanding that most women experienced period cramps.
 
Yeah but more painful than normal
 
Abnormally painful, I see.
 
In addition to the endometriosis, she also has weak bones @anakhro
Supercalifragilisic-osteoporosis
 
let me try
supercalifragilistic-commonflu
 
1:27 AM
@anakhro You know why she didn't learn about these problems for years?
Supercalifragilistic-underdiagnosis
 
Akiva, puns are the absolute lowest form of humour.
Despite this, my fiancee says you are funny.
"is a cool guy"
 
Yay
@anakhro You know what happened when she took the drugs prescribed for these?
Supercalifragilistic drug-induced psychosis
3
 
1:52 AM
huh
 
@anakhro: Tell your fiancée not to encourage DogAteMy. He's worse than Balarka.
 
did i return to the sites at a weird time
 
LOL ... it's always weird times.
Post your mathematics, @LittleBowsette.
 
@LukasHeger then what is the category equivalent to?
 
I have none I just was seeing what was going on after my forced year long hiatus
 
1:56 AM
Oh oh ...
hi @Leaky
 
hi
 
;-; siblings these days
i hope chat normally falls asleep a lot
 
2:33 AM
@TedShifrin I narrowly avoiding having to recount the entire lore of the math.SE chat with that relay.
 
 
2 hours later…
4:56 AM
@LeakyNun Yo
 
 
2 hours later…
6:47 AM
@Thorgott Hi
anyone here?
 
Hi there!
 
Hey
 
Hey guys
finally ! someone is here haha
 
Does it make sense in any way to call $1=0 x$ a singular equation in $x$?
I have a linear algebra case that appears to me of similar nature.
 
if pq two primes divides ab
it is the case that either p divides a or b right
same for q
 
6:56 AM
@Jacksoja yes
 
@Rudi_Birnbaum $0$ is a singular $1\times1$ matrix, so I guess that makes some sense in that context
 
well not "either"
 
is there a proof of this somewhere
I found euclids lemma when I searched
that is not it
 
it is precisely Euclid's lemma
 
let $a=pq$ and $b=pq$, then both divide both, so its not "either or" its simple "or".
which includes "and".
 
6:59 AM
you can also get it from the general equivalence of prime and irreducible elements in PIDs, but that's overkill
 
@Thorgott thanks!
 
and yes, Rudi raises an important point
 
that is one possible case
it could be a=p and b = q
in such case pq divides pq
but in my argument i want the general case
 
I was struggling a bit about this "singular" stuff. Since that $$ 0 x = 1$$ is intriduced as an "Ansatz" in a physical context. Then my intuitive impuls was to say its a non valid (non contradiction-free) Ansatz, and just to reject it. But they go through it until they discover that "singularity" an simply say one has to apply methods for the solution of singular equation systems and then observe that the solution does not have the full intended property.
 
so p divides either a or b
 
7:04 AM
And I find that wierd.
 
generally, the "either" is not true as has been pointed out
 
to be seen in a particular case that is not covered
thus its not general
 
Grüß Gott
 
Edward o..O
Servus?!
 
Ah jo! Namen geändert hehe
 
7:05 AM
Bischst du!
 
Genau I bin’s!
 
Wia gots?
@Thorgott any opinion on that?
 
Sitz gad im raucherbereich im Flughafen und gönn Ma an tschick während I ufn Flug wart
 
wohi?
(not good for you btw. ;-)
 
I fliag 2 wocha zruck noch England
Und jo I woaß aber egal hahaha
 
7:07 AM
Cool! Is des scho a bissle weita weg?
 
yeah, that seems like a rather weird Ansatz to me, but I have no clue whether or not that's sensible as far as physics are concerned
 
England moan i
@Thorgott You can extract some physics from it but for me ot was important to digest the logics first
 
England? 😁 jo scho, do hon i oan a halb Stund noch Frankfurt fahra müassa und denn fliegama über Amsterdam noch Bristol
 
I would formulate the buisness differently.
 
Goht nommol 2 stund glob i
Und denn fahrama 1.5 Stund vo Bristol noch Plymouth
 
7:09 AM
I also have another question about modular arithmatic
if we have a s.t gcd (a,m) > 1
 
Sind um 4 Uhr ufgstanda :(((
 
so it has no inverse modulo m
 
@EdwardEvans uff!
 
are some relations true ? like eulers theorem or fermat
for such elements
 
You need coprimality in Z/m for Euler‘s theorem
 
7:11 AM
I think the logics here depend on what is and isn't a valid Ansatz in physics and I'm a horrible judge of that
 
I see, but nothing could be said without that?
 
Wia wars Semschda
 
@Jacksoja as in, your element a needs to be prime to the modulus m for Euler‘s theorem
@Rudi was hoaßt semschda? Hahaha
 
@EdwardEvans okay thanks
 
semester?
 
7:12 AM
Ahhh tät Sinn macha
 
Well, there is general 3D vector field v and they make the Ansatz v=(0,a,b) and then they solve for a and b
Ja!
"Semeschda" my try at Alemannic...
 
Ja isch ganz okey gsi aber mine seelische gsundheit isch a kle schlecht also hon I mine prüfunga schiaba müassa
 
and then they say now we have a singular equation system and the solutions are not quite as we like
@EdwardEvans Ou tuet mr load
@Thorgott I think it should be the same as in maths
 
Hehe danke, as goht gad scho, aber jänner war ganz schlecht und I hon gär nix glearnat, was ganz schön kritisch isch wenn ma dPrüfunga am 6ta Februar hot hahaha
 
@EdwardEvans woll!
 
7:16 AM
Aber etz hon I no zit bis April also paaaaasst
 
@Thorgott would you call that a proper Ansatz in maths?
@EdwardEvans guet wos muesch-en macha?
 
it's an Ansatz that can't possibly yield a solution, so in maths, I would say no
but if, say, the equation is supposed to describe hypothetical physical properties of some object, who am I to judge
 
@Thorgott No nothing hypothetical all concrete and plain
Its also my feeling. I mean its possibly a matter of taste. But its definitely not my taste.
I now am not sure how I should handle this in my work, since its kind of a semi-established way already.
 
@Rudi bloß uf algebraische zahlatheorie und modulforma learna, wird scho goh hehe, und zwischadurch dua i mi uf a paar seminarvorträg vorbereita
 
@EdwardEvans Jo weard scho wern! Olles guete!!
 
7:20 AM
yeah, it wouldn't be mine either, but physicists sometimes do stuff that's pretty critical from a mathematical perspective
 
Wann got dr Fliagr?
 
Aber die kumman denn erscht im mai/Juni glob i
 
anyway, gotta go to uni now, bye!
 
Um 9.50 hehe
Tschau @Thorgott
 
@Thorgott thanks have a nice day!
 
7:22 AM
Und dia sind für lokale Klassenkörpertheorie und Darstellungen endlicher Gruppen :D
 
@EdwardEvans nice!
@EdwardEvans Haben Irreps was mit Fourierbasen zu tun?
 
Jaaaa wird cool :) für lokale klassenkörpertheorie goht min Vortrag übr den lokalen Kronecker-Weber Satz
Und des woaß I leidr nüd hahaha
Bei darstellungstheorie kenn I mi no gär ned us
Deswegat muass I mi druf vorbereita :D
 
Okey!
I muas iadz a loos ind uni!!
 
Oooo okay
 
Servus @Edward!
 
7:26 AM
An schöna i deam Fall!
 
Guetn Fluug!
merci!!
 
Pfiati :D danke
 
dir aa!
 
@EdwardEvans Hi I have a question
@TedShifrin Ted are you here?
 
7:51 AM
Ask it hehe
 
8:09 AM
@EdwardEvans
if we have that X^k = X mod rs , where r, s are primes ,but we only know their product, how can we find r and s ? k is such that gcd(k , (r-1) , (s-1) ) = 1
 
8:36 AM
any Assumption on X? I.e. is (X, rs) = 1?
 
@EdwardEvans no
but in that case
X is either r or s or rs so that case cause not problem
if the gcd >1
 
Well the (k, r-1, s-1) = 1 suggests Euler‘s theorem to me
 
it is ( k , (r-1) (s-1))
the product of those
the problem came from application of rsa in cryptography " possible failure of rsa)
suppose the attacker found out that for some known value X
that X^K = X mod rs
everything is known in that equation
the goal is to using that find r and s seperatly
k-1 has to divide (r-1) (s-1)
also rs divides m (m^(e-1) -1 )
from this since r and s are primes
we can conclude that r divides either m or (m^(e-1) -1 )
 
 
5 hours later…
1:32 PM
Morning, chat
 
Morning
 
2:24 PM
Happy Valentine's day chat
 
2:38 PM
Hello, how can I find the sufficient basis of any space, say $ R^4$ when two of its basis is already given in a formal way?
 
2:49 PM
Generally you can find a vector orthogonal to those two basis vectors, and then the vector orthogonal to those three vectors you already have
 
I have to ask a question about the operator $\nabla$ and the way it is used in taking the curl of magnetic field, anyone interested?
 
Just ask it
 
3:18 PM
ask!!
my interest is decreasing like 1/r^2
 
@geocalc33 :)
Biot Savart law says $$ \mathbf{B} (\mathbf{r}) = \frac {
\mu_0}
{4\pi}
\int \frac{
\mathbf{J} (\mathbf{r'}) \times \hat R }
{R^2} dV'$$
We need to be explicit over here : $$ \textrm{$\mathbf{B}$ is a function of x,y,z} \\
\textrm{$\mathbf{J}$ is a function of x', y', z'} \\
R = \mathbf{r} - \mathbf{r'} \\
dV' = dx'~dy'~dz'$$
Now, the operator $\nabla$ is w.r.t. unprimed variables.
taking the curl on both sides of the Biot-Savart Law and moving the operator inside the integrand we have $$ \nabla \times \left(\frac{ \mathbf{J} (\mathbf{r'}) \times \hat R } {R^2}\right)$$
We can expand this by the product rules and then we have $$ \mathbf{J} \left(
\nabla \cdot \frac{\hat R}{R^2} \right) - \left(\mathbf{J} \cdot \nabla \right) \frac{\hat R}{R^2}$$
Now comes the part which I'm unable to understand, my book writes for the second expression (that $\mathbf{J}\cdot \nabla ....$ one)
**Because the derivative acts only on $\frac{\hat R}{R^2}$, we can switch from $\nabla$ to $\nabla$' at the cost of minus sign** $$ - \left(
\mathbf{J} \cdot \nabla \right) \frac{\hat R}{R^2} = \left ( \mathbf{J} \cdot \nabla ' \right) \frac{\hat R}{R^2}$$
Please explain how we moved from $\nabla$ to $\nabla$'
If you want I can send the screenshots of the book.
 
3:40 PM
so when you use prime you mean that youre taking the derivative? @adeshmishra
 
@geocalc33 No, prime means the variables where the current flows, unprimed variables are the ones where we calculate the field.
 
Would anybody be willing to help me figure out if a question I have makes enough sense to post as a question?
 
4:22 PM
@AlessandroCodenotti I want to begin the study of Set Theory, right now I’m reading The Zakon Series on Mathematical Analysis
 
Question: Does the sum $\sum\limits_{i=1}^\infty (\frac{n}{i})^i$ converge for all $n$? It's been a while since I've done anything like this
 
@adeshmishra I don't know that series or how it's relevant, but I'm always happy if people want to learn set theory
 
But there are some proof which requires the knowledge of Logic, let's consider this one math.stackexchange.com/questions/3527913/…
 
I think so, but that's based on visualization of a plot of the points, and so is anything but rigorous
 
@AlessandroCodenotti I must tell you that I'm talking about the first course in Set Theory.
 
4:32 PM
I suppose I could do a squeeze. This should be less than $(\frac{1}{2})^i$ for all $i\geq 2n$ and I know $\sum\limits_{i=2n}^\infty (\frac{1}{2})^i$ exists
Or, rather, the absolute value should be less than that
So, I actually think that answers my question, unless I applied something incorrectly. (Now to find a form in terms of $n$ with no dependence on $i$ for the value the sum takes)
 
@adeshmishra One has to start somewhere
 
@AlessandroCodenotti If you have a little time then please check that link once and suggest me where should I begin from
 
4:53 PM
@AlessandroCodenotti Sorry for pinging you again, are you around ?
 
5:37 PM
@geocalc33 Please ping me when you get online.
 
@adeshmishra i'm online
 
@geocalc33 Can we discuss that?
 
@Rudi_Birnbaum ja definitiv
 
@adeshmishra sure, do you have the passage from the book I can look at?
 
Should I send the images or should I write it down?@geocalc33
In the footnote the book writes : **The point here is that $\mathbf{R}$ depends only on the difference between the coordinates; note that $$\left( \frac {\partial}{\partial x}\right) f(x-x’) = - \left (\frac{\partial}{\partial x’} \right)f(x-x’) $$
The book I’m using is Griffiths Introduction to Electrodyanmics, page number 234 (Pearson Publications)
 
6:15 PM
@adeshmishra okay
on the LHS you're taking the derivative with respect to $x$
 
7:13 PM
Quiet today
 
7:37 PM
Indeed
 
Took me a solid 5 seconds
 
 
3 hours later…
10:49 PM
certainly
 

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