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00:35
Problem: Let $f(X)$ be an irreducible polynomial over $F$ of degree $n$, and let $E$ be a field extension of $F$ with $|E : F | = m$. If $gcd(m,n)=1$, show that $f$ is irreducible over $E$.
Here's the solution in the back of the book:
Suppose you had a semigroup $S$ and you construct a new object out of it. Specifically, you add a single element $a$ and the rule $as=sa=a,\forall s\in S$, while keeping the other relations of the semigroup the same. I believe the resultant structure would still be a semigroup (correct me if I'm wrong), but is there a name for such a semigroup? Or would "semigroup union absorber" be the most appropriate?
Also, it's strange how we (people who are looking for questions to problems) come to the chat arout the same times.
I don't see why $|L : F| = mn$. I can show that $mn$ divides $|L:F|$, but that isn't enough...
I wonder if the solution is fallacious (it wouldn't be the first time). Going off that suspicion, I am trying to find a 3rd degree polynomial in $\Bbb{Q}[x]$ that is irreducible but is not irreducible $(\Bbb{Q}[\sqrt{2}][x]$...but I haven't had much luck.
01:08
First line tells you that [L:E] ≤ n, since L is a stem field for f_1 over E, and f_1 has degree ≤ n.
 
3 hours later…
03:41
Hey y9QQ
Hey, sorry it deselected your name as I was typing.
Thanks for volunteering to help
No worries! I like the question you posted -- I'm not entirely sure what the idea is to define the bijection.
I had the following idea, but I'm not convinced it actually makes any sense
If you have a map $\varphi: \pi_1 \to G$, then its kernel gives you a regular, path-connected cover $Y\to Z$ via the correspondence b/t subgroups and covers
So, H is the image of \varphi, maybe take a disjoint union of as many copies of $Y$ as there are cosets of $H$ in $G$.
So that would be the not-connected cover you get. But I'm not really sure that this is a $G$-cover. It's unclear to me how to define the right action.
Also, it doesn't vibe with Putman's example of Z\times G \to Z being the corresponding cover given the trivial map $\pi_1\to G$.
Apparently the action should be some sort of conjugation.
I've seen "monodromy" thrown around as well, but I don't know what that is, so I'm looking into it now.
03:50
Okay, let me take a look here ...
I'm not even sure how to map the fundamental group to the group of deck transformations. This is where the author uses monodromy.
So, you should think of monodromy and deck transformations as being kind of the same thing.
Let's say I have a cover Y\to Z, and some [a]\in\pi_1.
I can lift $a$ up to $Y$ using the cover.
Shouldn't the monodromy take a point in the fiber to the endpoint of a lifted path?
...which is the same point for loops.
Yeah, that's exactly it. You're getting a map to the symmtric group of the fiber above the base point
You have loops downstairs, but once you bring it upstairs it turns into a disjoint union of paths
In the proof this second paper gives, it isn't clear to me how to construct a cover given a homomorphism.
Are they disjoint paths only in the disconnected case?
03:57
Oh, they'll always be disjoint paths!
They'll lie on the path-connected space, but they'll be disjoint.
Well, I suppose disjoint outside of the endpoints
paths specifically? as in having distinct endpoints?
For example, take the degree 2 cover S^1\to S^1, and lift all of S^1
That loop downstairs lifts to two paths upstairs -- one on the top half of the circle, and one on the bottom half
Yeah, they share endpoints.
I mean a path γ(t) with γ(0)=/= γ(1)
Oh, sorry -- can you repeat your question then?
I was thinking of the monodromy action taking a point to the endpoint of the lifted path, but if a loop lifts to a loop, the endpoint is the same as the starting point. In that case, the action seems sort of trivial, unless the paths are specifically not loops
04:05
Ah, right. So the monodromy action is defined as taking a based loop downstairs, in Z, and then lifting that loop up into Y. When you do that lifting, you get paths that run between the points in the fiber above the basepoint. Now, pick a particular point, p, in that fiber. Then there is a unique lift of your based loop downstairs to a path upstairs that starts at p. If q is the endpoint, then the monodromy action is the action taking p to q.
You're absolutely correct, the paths are only loops when the cover is degree 1 -- i.e. the trivial cover.
This gives a well defined map from \pi_1 of Z to the symmetric group of the fiber above the basepoint. If my head is on straight, I believe that the image of that map is the deck transformation group.
Oh, so p=/=q if the cover has more than one sheet?
Right! Because p and q are in the fiber above the basepoint, which will have however many points as the sheetedness of the cover
The reference for this discussion in Hatcher is around page 70, I believe.
I think you mentioned you were studying from that book
Oh wow that actually makes a lot of sense
I'm thinking R->S^1
Yeah, exactly!
That's the perfect example.
That's so cool
04:11
Right? Covering space theory feels like a lot of magic to me
so in this example, I imagine the monodromy action to be rotating the helix above the circle.
All the way to another point in the same fiber
Right, which is exactly what the deck group \mathbb{Z} is doing to the helix.
Same thing.
OH, and it's a kind of just a permutation of the sheets.
Yeah that is magic
04:14
So, take a look at the proof of the bijection in this van_Beek paper
Given a G-cover, this is exactly how they build a map from $\pi_1(X,x)$ to $G$
Unfortunately, they aren't really giving all the specifics ...
In this case the author mods out by isomorphism classes of G-coverings. That is equivalent to restricting to regular g-coverings?
They're modding out by isomorphism of G-isomorphisms
So, to give an example of what this means, say you have a G-cover Y of a space Z.
So identifying all covers with isomorphic automorphism groups?
This really means that there is an action of G on Y that when you look at the quotient of Y by the orbits of G, the natural quotient map gives a cover of Z
But, I could define a "different" action by first doing some automorphism of G before I let my elements of G act on Y.
It's basically the same action, but technically different because the action (the map Y\times G \to Y) is a different function.
This "modding out" is the author just saying they don't want to consider that sort of situation.
That's how I'm understanding it, anyway.
You might appreciate these notes: math.jhu.edu/~jmb/note/gsetcov.pdf
04:30
Yes, these are extremely helpful, thank you.
My group theory is a bit rusty, which would explain my trouble here.
Hm. So I think my thought above about the isomorphism bit is a little wrong. I'm looking at Fulton's book (which Putman references as the source for the proof you're looking at) to get a sense of what a G-cover actually is, and he defines what the isomorphism is.
You might want to find a copy of Algebraic Topology: A First Course by W. Fulton if you can, and flip to page 160
I just picked up a copy of Fulton, but he seems to use a bunch of notation that Hatcher doesn't use.
I find I'm having to backtrack to work out notational conventions
This particular section about G-covers seems to be fairly notation light.
Ah yes, reading now
Also, thank you for helping me out. I hope I'm not keeping you up too late.
Note the example Putman gives for the trivial homomorphism in the 1-1 correspondence and the trivial G-covering.
Oh, of course not! For some reason my sleep schedule has been all sorts of weird lately, so this isn't quite getting late for me.
Plus this sort of stuff is what I love, so I'm having fun finding all the holes in my understanding here
Plus I've been there -- self studying algebraic topology is a bit of a nightmare.
04:44
I'm seeing Galois a lot. Perhaps I should have taken that topic more slowly...
Mm, where are you seeing it?
Right, okay. Yeah, it turns out that number theory and algebraic topology and lots of other areas of math display the same behavior in different areas
The "Galois cover" that link mentions is exactly the normal cover/regular cover you know, and the "Galois group" of the cover is just the deck transformation group. The author is just trying to be heavy handed with the relationship between covering space theory and number theory.
Actually, if you're finding that your group theory chops are not up to snuff, you might want to consider taking a look at Algebra: Chapter 0 by Aluffi. It goes through group theory through the lens of category theory, which is the viewpoint a lot of your sources are taking.
This may be a stupid question, but Fulton says there is a projection map p:Y->Y/G sending each point to the orbit containing it. That must be injective, right?
Only if G acts trivially on Y!
Any point in the same orbit will go to the same point in Y/G
04:55
Ah, not injective. I dont think I had thought that one through. I was thinking that a point can easily be in the orbit of two points. Therefore it can be mapped to two orbits, but those orbits must be the same.
I was thinking one-to-many instead of many-to-one
But I just realized that if it's in the orbits of two points, those orbits are actually the same
since G is a group
Yup! Exactly.
05:16
@y9QQ Okay, I'm going to head off to bed now. I hope our discussions were helpful!
Yes, they were!
Thank you again.
I really enjoy this sort of thing, so if you'd like to reach out again, you can just email me.
I think I have my website linked on my math.se account.
I'm working through Fulton's exercises now.
I may do just that, if you don't mind.
Oh, please do. Seriously. I usually have a group or two of undergraduates that I do reading courses with each semester because I like working through stuff like this.
I'
I'm a physicist, but I find myself tending toward this sort of material much more lately. I'm lucky to have the support of stackexchange. Thank you again!
05:25
Glad to be of some help. Good night!
Good night!
 
2 hours later…
07:49
Does the independency of continuum hypothesis prevents me of finding a set of real numbers which size is strictly in between of N and R ? (Naturals and reals)
Depends on what you mean by "finding"
08:29
@AlessandroCodenotti Finding like defining set A such that requirement holds.
Dont know how else to put it
Does the independency say that such set might exist but it would be impossible to find/define it ?
What Cohen did to show that $\neg\mathsf{CH}$ is consistent with $\mathsf{ZFC}$ was to start with a model in which $\mathsf{CH}$ holds, then use forcing to add $\aleph_2$ new reals and build a new model $V[G]$ in which $|\Bbb R|=\aleph_2$. In $V[G]$ the set of reals of $V$ has cardinality $\aleph_1$, does this count as "defining" a set of intermediate cardinality?
For a simpler example pick a well ordering of $\Bbb R$ (which exists by AC) and consider the set of all reals which have only countably many predecessors in this ordering. This set has cardinality $\aleph_1$ so it is a set of intermediate cardinality if CH fails
09:31
Interesting
09:41
@AlessandroCodenotti what is $\aleph_1$ and how would that be intermediate if CH fails ?
$\aleph_1$ is the least uncountable cardinal, or the smallest cardinal above $\aleph_0$. It would be intermediate because CH is the same as $|\Bbb R|>\aleph_1$
10:15
I don't see why $|L : F| = mn$. I can show that $mn$ divides $|L:F|$, but that isn't enough...
 
2 hours later…
12:13
Hello, can someone help me with a statistical modelling question
https://math.stackexchange.com/questions/3184572/statistical-modelling-question
13:07
@JCheong I have always found cross validate very helpful for stats questions. They have very deep knowledge there about statistical inference and econometrics etc.
 
1 hour later…
14:32
In a cellular automaton, a finite pattern is called a sawtooth if its population grows without bound but does not tend to infinity. In other words, a sawtooth is a pattern with population that reaches new heights infinitely often, but also infinitely often drops below some fixed value. Their name comes from the fact that their plot of population versus generation number looks roughly like an ever-increasing sawtooth wave. == In rules with small replicators == For instance, in Rule 90, a one-dimensional elementary cellular automaton, the population size starting from a single live cell follows Gould...
Can I get someone to check me on something? If I have a vector U [1,2] and vector V [3,2], the Hademard product would be [3,4], not [3,6], yes?
@JohnP LGTM
@LeakyNun Have you looked at the wiki page for it? I haven't had math in 15 years. A simple yes or no would have helped. I'm working through a data science course for python, and the example doesn't match what I think it should be.
Definition 1.1.Let A and B be m×n matrices with entries in C. The Hadamard product of A and B is defined by [A◦B]ij= [A]ij[B]ij for all 1≤i≤m,1≤j≤n.
If you don't understand the notation, that makes no sense.
I understand the notation
yes it is [3,4] not [3,6]
Thank you, I appreciate that. After looking at this module, I'm thinking I may need remedial courses in linear algebra as well.
[A◦B]ij= [A]ij[B]ij for all 1≤i≤m,1≤j≤n. - What is that saying, if you don't mind?
14:57
My combinatorics is bad. Say I have $N$ objects that are either of type $A$, or of type $B$. I know $N_A$ and $N_B$, the number of objects of each type. I know that $\frac{1}{2}$ of all objects come in exclusive binary pairs (of mixed or same type). I want to find the approximate number of objects of type $B$ which are paired with other objects of type $B$.
Hi @Mathphile
is there any function that covers the entire xy coordinate plane
What do you mean by "covers" and what is the domain and codomain of your function?
by covers i kind of mean like fills up the entire area of the plane
15:02
And of the domain and codomain?
i dont know how to explain this in technical terms
i know im probably sounding stupid by now
Well a function is something that takes an input x and spits out an output f(x).
So where are you taking your input from?
Real numbers? The plane itself?
An interval?
is a measurable function of a local martingale always another local martingale?
well i think the function i am trying to describe would be like n(sin(nx)), n tending to infinty
I am guessing you are wanting space-filling curves.
15:06
cuz that woold cover then entire plance
But I can't be sure.
yes
space filling
But this is a limiting procedure.
In mathematical analysis, a space-filling curve is a curve whose range contains the entire 2-dimensional unit square (or more generally an n-dimensional unit hypercube). Because Giuseppe Peano (1858–1932) was the first to discover one, space-filling curves in the 2-dimensional plane are sometimes called Peano curves, but that phrase also refers to the Peano curve, the specific example of a space-filling curve found by Peano. == Definition == Intuitively, a continuous curve in 2 or 3 (or higher) dimensions can be thought of as the path of a continuously moving point. To eliminate the inhe...
This page predominantly concerns filling the unit square.
Since plane curves are usually from [0,1] to the plane, they will never be able to "fill" the entire plane.
I am interesting on this curve, but curves like peano is not a function..
However you can do it for (0,1) I think.
15:08
this link is interesting
It seems there is a function whose Hausdorff dimension is 2..
@LarryEppes why are they not functions?
but i guess someting like n(sin(nx)), n tending to infinty would work too
ehh, i think the question is find a function of variation on \RR. and only one y corresponding to x.
@LarryEppes he never said anything about that.
Oh I misread what you said.
Peano curves are single-valued.
They are not injective in general, though. But that's not what you said either.
Also, I don't really think the limit of n(sin(nx)) is really counted as a space filling curve.
Because its limit doesn't exist.
That's more so an issue that you are taking it as a function R-->R though.
15:24
On a class of fractal functions with graph Hausdorff dimension 2
this is my previous supervisor's article... all I know about dimensions of a function
Any plane filling curve has Hausdorff dimension 2, @LarryEppes
yes, so if these curves with Hausdorff dimension2 can filling the plan? I donnot know...
15:43
@LarryEppes are you a graduate student?
I love mathematics!
@J.Doe what kind of math?
Now that's a deep question
I'm still a high schooler sniffles
Well what kind of math do you enjoy now?
 
1 hour later…
17:04
**My attempt:-**

$$\int_{0}^{1}f(x)dx=\sum_{k\in \mathbb N}(\frac{2^k-1}{2^k})*(\frac{1}{2^k})=\sum_{k\in \mathbb N}(\frac{1}{2^k}-\frac{1}{4^k})=1-\frac{1}{3}=\frac{2}{3}$$
Did I just see "Riemann-integrable" and actually computing an integral in the same question
@LeakyNun did you not get the fun exercise of doing Darboux sums by hand for special functions?
oh no
"fun"
It's actually quite neat.
To be able to do something formal like that by hand.
Takes a clever partitioning, and the likes.
Very neat.
17:22
does anyone know how to use mathematica here?
well lets just say the guys there weren't that friendly to me
actually i need help to plot a infinite series
I see your question.
The reason they were not so nice is because you demanded they upload a picture for you.
and i need something which is a lot more complicated for desmos to handle
yes
and i completely understand them not being nice
You should instead ask them for advice on how to plot it so you can plot it yourself.
But they have a good point that you should maybe ask that sort of question on MSE instead.
17:26
well i am out for a few days and dont have access to a computer with mathematica with it
only have been using my phone to type here
plus i dont know much about mathematica
Well they gave you some help despite that.
They referred you to the plot function.
Maths department asked me to do a video proof of the irrationality of $\sqrt{2}$. Was stood at the board ready to start, the film crew are there waiting for me to start and the first thing that comes out of my mouth is "We'd like to show that $\sqrt{2}$ is rational"
@ÍgjøgnumMeg lol
painful
haha
@ÍgjøgnumMeg we all know it's 7/5
17:30
Exactly
Please help me with this proof that, There is no continuous bijection between $\Bbb R^2$ and $\Bbb R$: Am I getting it correct: Suppose there is continuous bijection $f$, then $\Bbb R^2-\{0\}$ is continuous surjection $\Bbb R-\{f(0)\}$, but image of connected set under continuous map connected, hence contradiction.
Am i getting it right?
Basically.
If there were a bijection, then connectedness after removing any single point would be preserved.
Is there a fast way to find the tangent for an ellipse?
if we have 4 random variables A, B, C, D and we know that A+B and C+D are independent. Does that guarantee that A is independent of C and D?
@Arthur B can be -A for all we know
17:36
@anakhro you mean preserved in the domain? ok, thank u
@Silent preserved as a topological property.
Please elaborate this last sentence!
What part of it doesn't quite make sense?
@LeakyNun Ah ok. So B = -A implies A+B is constant and is therefore independent of C+D, thanks
that 'preserved as a topological property.' is i can't get
17:40
lmao
@LeakyNun so are you thinking A = C = -B as an example where A and C are dependent but A+B and C+D are independent?
that would work
cool, thanks!
@Silent you will have to give me more than that. Do you know what a topological property is?
yes. A property that is invarient under homeomorphism.
17:48
So is P(X) = "for all x in X, X\{x} is connected" a topological property?
yes it is. Thank you very much
if a finite measure on R has compact support, is any bounded measurable function integrable?
I probably need more hypothesis on the measure
18:28
T/F: For any $n ≥ 2$ there is an $n×n$ matrix $A$ with real entries such that $A^2 = A$ and trace$(A) =n + 1$.
My try: for matrix with $A^2 = A$, eigenvalues are $0,1,-1$, hence at best we can have n eigenvalues all of which 1, $1+1+\cdots+1=n$, hence above statement false. Am i right?
idempotent matrices have eigenvalues 0 or 1.
No -1
Otherwise it would fail to be idempotent.
$A^2v = A(-v) = -Av = v$
when $Av=-v$
Oh! Thank you very much
@LeakyNun hmm this looks true to me in first glance
19:32
1
Q: Theorem 1.27 in Rudin's Functional Analysis

user193319 1.27 Theorem: Suppose $Y$ is a subspace of a topological vector space $X$, and $Y$ is an $F$-space (in the topology inherited from $X$). Then $Y$ is a closed subspace of $X$. Here is Rudin's proof: I don't see why $y_0$ being in $\overline{x + W}$ for every open nbhd $W$ of $0$ implie...

20:25
A bilinear form is the same to bilinear function?
Can someone check my proofs? Please? Is about linear algebra
20:48
Hi all!
i have a question. is there any clear connection between inscribing and covering of geometric objects?
I am thinking about something along the line of duality.
21:09
circumscribe is more of a dual than covering
21:26
Basic trig: I am rolling a ball down a ruler
If I want to find the height of the ball from a certain horizontal distance on the ruler, do I need to multiply the distance by the sine or by the cosine of the angle from the ruler from the ground?
22:14
@anakhro i think this is what i mean. putting a closed curve in a say square s. t. they touch at least in each edge but do not intersect. is that circumscribe?
Because i find it easy to show that any closed curve can be circumscribed by a square. While what dual wmmm
Now I thought that might be a way to the Töplitz conjecture.
Sorry for the last but one sentence, I'm using a phone.
 
1 hour later…
23:36
Right now I'm looking to learn more about martingales.... in particular I have a situation where there is a continuous martingale y(t) which can take on values in the interval [0,1] inclusive. But, the t variable doesn't really mean anything. So I'm looking for statistical properties that are completely independent of any transformations that could be applied to the t variable.
Not sure if this is a question on-topic for the main site.
^ here's an example of such a process. Let y(t) be the probability that a certain team will win a basketball game, as a function of time through the game.
An an example of a property that I'm talking about: Given that team A currently has a 75% percent chance of winning, what's the probability that the chance of winning will ever drop below 60%? (The answer being 0.625)
But the main motivation is... given a chart like the one above, is it possible the determine whether or not it is properly calibrated, and if it's not calibrated, then calibrate it?
"Calibrated" meaning that it actually is a martingale process, and "calibration" meaning applying some transformation to the y-values to make it a martingale process.

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