2:22 AM
I have a copy of "The VNR Concise Encyclopedia of Mathematics" from 1977 - I found a copy in a room for a teacher I was subbing for and liked it so much that I got one on Amazon (just a few bucks if I recall) - is there a better text that covers the same material?

3:18 AM
Dec 23 '19 at 1:15, by Mike Miller
yeah, right.

Hey chat
Bringing back a question I've asked: how do I prove the primitive element theorem for separable extensions?
It looks like the finite case requires a bit of abelian groups (I didn't study them with depth but I will), but the nonzero char case for infinite fields scares me, lol

3:41 AM
hi @Ted!

3:53 AM
@EdwardEvans which is the best method to solve ..Do I do componendo-divideno or cross-multiply or what?

3 hours later…
6:59 AM
Can a polynomial fit any curve?

3 hours later…
10:01 AM
Does fixed point exist for continuous function in interval [0,1] union [2,3]?

@mathsstudent try 3-x

f(X) =3-x

Anyone here familiar with Rabin Miller prime test?

@mathsstudent try x+1

1 hour later…
11:36 AM
Quiet huh
Let $M$ be a manifold with a smooth foliation $\mathscr{F}$ and a 1-parameter family of measures $\{\mu_{\theta} : \theta \in \Bbb R\}$ (continuous in $\theta$ in the weak sense). Suppose $\mu_{\theta}$ restricted to the leaves of $\mathscr{F}$ are unchanging as $\theta$ varies, so the family is $\mathscr{F}$-invariant in that sense
I guess you would say on the flowbox coordinates $\mu_{\theta} = \mu \times \nu_{\theta}$ where $\mu$ is a fixed measures on the leaf coordinates and $\nu_{\theta}$ is some $\theta$-dependent measure on the transversal coordinates
So the family $\{\mu_{\theta}\}$ should be thought as an evolution on $M$ where the leaves are given some fixed measures each which do not evolve in time and the way the measure on all of $M$ evolves is if the leaves come togather to concentrate or move away to dissipate
I will call this a sufficient foliation for the family of measures.
I will call a sufficient foliation $\mathscr{F}$ for a family $\{\mu_\theta\}$ to be a "minimal sufficient foliation" if for every other sufficient foliation $\mathscr{F}'$ for $\{\mu_\theta\}$, $\mathscr{F}$ contains $\mathscr{F}'$, which is to say the leaves of $\mathscr{F}$ contain the leaves of $\mathscr{F}'$.
I'm having trouble imagining what an ancillary foliation should be. I expect it's a foliation $\mathscr{F}$ on $M$ such that for a foliated open subset of $M$ (an open subset which is a union of leaves), if you evaluate $\mu_\theta$ on it, then the value doesn't change wrt $\theta$. So the evolution leaves $\mathscr{F}$ totally invariant, the leaves don't concentrate or dissipate wrt each other as $\theta$ varies

12:15 PM
@BalarkaSen let's play some Ke2

Not really feeling up to it

1 hour later…
1:20 PM
Hi guys
Between Tu and Lee which one's a better intro to manifolds/differential geomemtry

1:34 PM
there's Lee--Tu difference

1:57 PM
I like Lee

what's the difference @LeakyNun
@Slereah why you prefer lee more than tu ?

@Lelouch it's a pun

Personal preferences?
I dunno

i mean can you elaborate a bit ?
like what aspects of lee you like and tu you don't like ?

I figured it's supposed to be a pun, but I don't get it

2:12 PM
@Thorgott little
in my Asian accent

oh lol
that's a bit of a stretch

so my accent merges "i" into "ee"

Tu is better
I liked the pun

and "tle" is pronounced like "too" or "tow"
@BalarkaSen takes off hat

2:37 PM
hi , is it useful to figure out that N is prime only if the parametric equation $\left(\sqrt{N}\cosh\left(t\right),\sqrt{N}\sinh\left(t\right)\right)$ has only one integer solution for t in Real
real positive and does not work for 2
(N is a odd number)

2:57 PM
13 hours ago, by Aaron Hall
I have a copy of "The VNR Concise Encyclopedia of Mathematics" from 1977 - I found a copy in a room for a teacher I was subbing for and liked it so much that I got one on Amazon (just a few bucks if I recall) - is there a better text that covers the same material?

3:27 PM
@AaronHall you can try The Concise Oxford Dictionary of Mathematics

@rapasite that seems to be just a straight up dictionary
Everything is organized alphabetically, but the VNR is organized by topic, each thing building on the previous

3:43 PM
Hi chat! In Spivak's Calc on Manifolds he writes in his proof of the uniqueness of the derivative of $f:\mathbb{R}^n\to\mathbb{R}^m$:

If $x\in\mathbb{R}^n$, then $tx\to 0$ as $t\to 0$. Hence for $x\neq 0$ we have $$0=\lim_{t\to 0}\dfrac{|\lambda (tx)-\mu(tx)|}{|tx|}=\dfrac{|\lambda(x)-\mu(x)|}{|x|}$$

But I just don't see how that very last equality follows

Are $\lambda$ and $\mu$ linear?

@Thorgott Do you know some statistics by any chance

4:13 PM
not more than what's covered in a standard course on probability theory

@Thorgott ooh that's why! I completely missed that for some reason ^^ thanks again!

4:30 PM
I'm afraid I don't know about these

Hi @BalarkaSen
@BalarkaSen can you please elaborate why ?

5:36 PM
Belated hi, a @Balarka.

5:58 PM
I need to proof that the determenant of a a matrix which happens to be symmetrical is zero.
is there a clever way to calculate this. i have been trying to figure out some shortcut since doing it with leibniz formel is too long.
We have not done Eigenvalues for matrices yet so thats out of the question
is a 6x6 matrix

Is it obvious for other reasons that the rank is less than $6$?

I do not think so if you mean tehres a zero coloumn or row thats zero
however the diagonal is zero

Well, I didn't mean that obvious :)

By the way. Hello mr Shifrin.

In what context are you getting this assignment?

6:02 PM
This matrix is symmetrical with a diagonal of zeros.. I have calculated this matrix myeslf. Each element of the matrix is equal to the number of inversions of composotion of two Permutations.
Its a part question of a bigger question.
I do not think you can figure out what i am talking about without doing the previous part-questions leading up to this one. But i ask in general if theres such a rule for symmetrical matrices, or those whose diagonal is zero.

No.
Let's see the matrix.

$\begin{array}{rrr} 0 & 1 & 1 & 2 & 2 & 3 \\ 1 & 0 & 2 & 1 & 3 & 2 \\ 1 & 2 & 0 & 3 & 1 & 2 \\ 2 & 1 & 3 & 0 & 2 & 1 \\ 2 & 3 & 1 & 2 & 0 & 1 \\ 3 & 2 & 2 & 1 & 1 & 0 \\ \end{array}$

It may be that the way you constructed the matrix tells you that the matrix has to be singular. But it shouldn't be too hard to put the matrix in echelon form, or to go far enough to see you get at least one row of zeroes.

6:26 PM
@TedShifrin have you ever considered contributing to the beast academy?

Don't know what you talkin' about.

The Art of Problem Solving for younger students

Oh, right, part of AoPS. I did see stuff about that when I was there.

@TedShifrin Thats what i thought. I just didnt figure it out.. well i will just brute my way thru

6:49 PM
Hey chat

Hi Lucas.

@MadSpaceMemer the permutation definition of the determinant tells that youâ€™ll always have a zero in each term of the sum
Since every column has at least one zero

@Lucas: That's nonsense.

Yeah, thatâ€™s nonsense :p
It seems I canâ€™t delete the message, tho

OK. :)
You don't get much time to think around chat.
Perhaps he'd rather you stopped by his office during office hours?

6:56 PM
@Ted, as a professor: itâ€™s been a week or two since my advisor last replied me. I need help with a proof (primitive element theorem), but he didnâ€™t reply and thatâ€™s unusual. Whatâ€™s more probable: did he miss my email or is he on vacation?

Is this vacation time?

Yeah, but he told me itâ€™s ok to ask for help.

Well, I'd suggest you make some progress and slightly change your email, and then say "I don't know if you missed my email a week ago, but here's where I've gotten, but I'm still stuck ..."

And Iâ€™ve sent plenty of emails since our classes ended
Gotcha. Thank you, Ted :)

What about the primitive element theorem?

7:09 PM
Thorgott, maybe ping Lucas in case he's not paying detention?

Iâ€™m back

So: Langâ€™s book is confusing. He proves results for characteristic 0, then for subfields of the complex numbers and suddenly goes back to arbitrary fields
Every extension of characteristic 0 is infinite and separable, so the primitive element is easy.
Oh, and for subfields of the complex numbers, he proves that there are exactly $n$ embeddings for a extension of degree $n$
Thatâ€™s a result he uses to prove the primitive element theorem for subfields of the complex numbers
With all this partial info, I want a general proof of the primitive element theorem for separable extensions

Now you should be pinging @Thorgott. :D

So I can proceed with the fundamental theorem of Galois theory as weâ€™ve talked before, @Thorgott
Iâ€™m also wondering if there are analogous results about the number of embeddings of an arbitrary extension into its algebraic closure
Maybe itâ€™s $\leq n$?

7:27 PM
The number of embeddings into an algebraic closure (which is independent of the choice of closure) is called the "separable degree" of an extension. It is always $\le n$ and the extensions where it's $=n$ are precisely the separable extensions (this is the definition, usually).
That said, I'm not sure what proof Lang gives, but I think there's no need to make a distinction by characteristic to prove the primitive element theorem.
The finite and infinite cases are usually treated separately though
I'm not sure how you want to use the fundamental theorem of Galois theory. The proof of the fundamental theorem that I know requires the primitive element theorem to begin with.

Iâ€™m not trying to use the fundamental theorem to prove primitive element theorem, but the reverse. Could you link me to a proof of the general case?

You can take a look at these notes from Keith Conrad
There's a fun alternative way of proving the primitive element theorem for extensions of finite fields by explicitly counting the number of irreducible normed polynomials of given degree

@Lelouch Lee spends too much time on analytic details and doesn't talk about manifold topology/geometry. His smooth manifolds book don't deal with transversality, normal neighborhoods, etc and his Riemannian manifolds book ends abruptly after defining curvature as far as I remember.

Which leads to a prime number theorem for irreducible normed polynomials over finite fields

Tu's smooth manifolds book is a good intro to manifolds and his differential geometry text talks about bundle geometry, moving frames, etc.
Belated rehi @TedShifrin
@Lucas The standard proof of primitive element theorem by showing $F(\alpha, \beta) = F(\alpha + c\beta)$ for separable elements $\alpha, \beta$, by choosing $c$ appropriately, works out if $F$ is an infinite field.
For finite fields you have to deal with it separately, but it's nonetheless easy as finite extensions of finite fields are easy to classify
The reason it has the potential to fail for finite fields is because you want to avoid $c$ to be $(\alpha_i - \alpha_j)/(\beta_j - \beta_i)$ where $\alpha_i$, $\beta_i$ are the conjugates of $\alpha, \beta$. These constitute finitely many choices; what if it exhausted every element of $F$?

7:52 PM

Thall

scrambling around my bed to find earphones

Can't listen right now but I'll check it out
Have you heard the new Vildhjarta single? @Edward

yeah it's great

Indeed, I'm looking forward to the new album

8:06 PM
Aaaaaye
Thallbum
Fractalize have cool ass rhythms

i am missing out here

Vildhjarta are great

lol i love to partake of djent every now and then

recommend me some good dsbm @Edward

oh man
know this?

8:10 PM
Nope! Gonna listen in a bit

this is cool hahaha

1 hour later…
9:22 PM
@Lucas the standard proof for the primitive element theorem is the one that @Balarka mentioned. There's another proof. By Galois theory (passing to the normal closure of the extension), a finite separable extension has only finitely many subextensions. Now if the ground field is infinite, then a theorem in linear algebra says that a vector space is not the union of finitely many proper subspaces.
Thus one can choose any element not contained in any proper subextension and this will automatically be a primitive element.
you need to deal with finite fields separately here as well
the standard proof is probably better because it needs less

Can you establish enough Galois theory to prove that without having the primitive element theorem?

yes, absolutely
you don't need the primitive element theorem at all to develop Galois theory
I'm not saying you can't utilize it somehow, but modern treatments following Artin usually don't need it

Hello all! I am trying to prove in set theory: $$P(A-B)\subseteq(P(A)-P(B))\cup\{\emptyset\}$$ ($P$ indicates for Power set)
My proof is:
$$X\in P(A-B)\implies X\in P(A\cap B')\implies X\subseteq A\cap B'\implies X\subseteq A\wedge X\subseteq B'\implies$$
$$X\in P(A)\wedge X\in (P(B))'\underbrace{\implies}_{p\to p\vee q}(X\in P(A)\wedge X\in (P(B))')\vee X\in\{\emptyset\}$$
$$\implies(X\in P(A)\cap (P(B))')\vee X\in\{\emptyset\}\implies X\in P(A)-P(B)\vee X\in\{\emptyset\}\implies X\in(P(A)-P(B))\cup\{\emptyset\}.$$
What I am not sure is in the part: $X\subseteq B'\implies X\in(P(B))'$. I think it should be $X\in P(B')$, but if the last is correct, then my proof is wrong. What do you think? Thank you!!

9:43 PM
Hmm, I think if you want to prove Galois correspondence, you at the very least want to know that if $K$ is a field and $G\subset\mathrm{Aut}(K)$, then $K/K^G$ is a Galois extension with Galois group $G$. The proofs I've seen of this use the primitive element theorem. I just had a look at Artin and he uses it too.

possibly dim question.. how many different binary strings have hamming distance <=k from a binary string of length n?

9:59 PM
@manooooh that's wrong as it stands: what if $X=\emptyset$?

@Thorgott hmm, you're right, I retract what I said. The proof I gave can still be made to work if you prove that a finite separable extension has only finitely many subextension without Galois theory, e.g. by using tensor products of algebras
if $E/K$ is separable with $[E:K]=n$ and $\overline{K}$ is an algebraic closure, then $\overline{K} \otimes_K E \cong \overline{K}^n$ (as algebras). So if $E/L/K$ is a subextension with $[L:K]=m$, then tensoring the inclusion $L \hookrightarrow E$ gives an injection of $\overline{K}$-algebras $\overline{K} \otimes_K L \hookrightarrow \overline{K} \otimes_K E$, but since $\overline{K} \otimes_K L \cong \overline{K}^m$, it is a subalgebra generated by idempotents,
but there are only finitely many idempotents in $\overline{K} \otimes_K E \cong \overline{K}^n$, all components have to be $1$ or $0$

10:23 PM
@Thorgott I don't know what do you mean. Could you explain please?

I'm not very fit with tensors. Why does it follow that it's generated by idempotents?
@manooooh you are not sure about whether $X\subseteq B^{\prime}$ implies $X\in(P(B))^{\prime}$. I'm asking what happens if $X=\emptyset$.

@Thorgott we have $\emptyset\subseteq B'\to\emptyset\in(P(B))'$. The antecedent is obviously true; the consequent no because we have to know what $B$ is

10:39 PM
you don't have to know $B$ is, the consequent is always wrong

@Thorgott I don't think so. Can you find a counterexample?
I couldn't; that's why I supposed it was true
Ohh you are right, thank you!
So my proof is wrong. Is there another proof?

@Thorgott because $\overline{K}^m$ is generated by the idempotents $(0,\dots, 0,1,0,\dots, 0)$

10:54 PM
oh, I was missing the forest for the trees
there are only finitely many copies of $\overline{K}^m$ in $\overline{K}^n$

right, that's the argument

How can we ensure that there are no two intermediate extensions $L,M$ with $\overline{K}\otimes_KL\cong\overline{K}\otimes_KM$? Does the tensor allow cancellation here?

yes, $\overline{K}$ is faithfully flat over $K$ which means this kind of cancellation is allowed

@manooooh a careful examination will show that the empty set is the only set for which your proposition fails

This more tensor-heavy approach to separable extension comes up naturally in alg geo or if you do Galois theory à la Grothendieck
actually it can happen that there are two intermediate extension with $\overline{K} \otimes_K L \cong \overline{K} \otimes_K M$, but they will still have a different image in $\overline{K} \otimes_K E$
that's just linear algebra: if two subspaces are the same after extending the base field, the subspaces have been the same to begin with

11:02 PM
ooh, I see
that's neat

@Thorgott but $P(\emptyset)\subseteq(P(\emptyset)-P(\emptyset))\cup\{\emptyset\}$

11:20 PM
yes, what I'm talking about is the intermediate proposition, in case that wasn't clear

@Thorgott oh, in the "proof". I see. Thank you. Do you have a real proof? I couldn't find it

well, your proof is fixed once you fix that intermediate step
what I'm saying is that the $X=\emptyset$ counterexample is the only one, so if $X\subseteq B^c$, then $X\in\mathcal{P}(B)^c$ or $X=\emptyset$

@Thorgott I can't fix the intermediate step

that also explains why the emptyset is treated separately in the conclusion (in your proof, you just add it tautologically, which suggests it could actually be omitted from the conclusion, but it can't)
i just told you how to fix it

Yes, the RHS of the union was quite unusual
Oh I see what you mean
In that case we have: $$X\subseteq A\wedge X\subseteq B'\to X\in P(A)\wedge(X\in(P(B))'\vee X=\emptyset)\to(X\in P(A)\wedge X\in(P(B))')\vee(X\in P(A)\wedge X=\emptyset)$$
I don't see why $X\in P(A)\wedge X=\emptyset$ is equal to $\{\emptyset\}$

11:34 PM
it's not equal $\{\emptyset\}$ and it shouldn't be, you're manipulating logical statements, not sets
remind yourself of what you want to conclude

@Thorgott ok, so I should have write $X\in\{\emptyset\}$ no $\{\emptyset\}$
@Thorgott I want to conclude $X\in (P(A)-P(B))\cup\{\emptyset\}$

Hey guys, I was wondering if it is possible to know the inverse image of two subsets, i.e, $f^{-1}(\{A,B\})$.

11:50 PM
@AlekMurt: You can't write $\{A,B\}$ when $A$ and $B$ are subsets. At any rate, your question is totally vague. What $f$ are you talking about, for starters?

@TedShifrin any thoughts on my reasonings?

What reasonings?
Way too much stuff to read.

I have understood that $X\subseteq A\wedge X\subseteq B'\to X\in P(A)\wedge(X\in(P(B))'\vee X=\emptyset)\to(X\in P(A)\wedge X\in(P(B))')\vee(X\in P(A)\wedge X=\emptyset)$
I want to prove: $P(A-B)\subseteq(P(A)-P(B))\cup\{\emptyset\}$

You sure love this symbolic garbage, don't you?
I write math in words and sentences.

Yes, I love to put parentheses and so on

11:52 PM
Real mathematics is not done in long lines of symbols. I hope you get over this very fast.

bonsoir @Ted

Guten Abend, Herrn @Lukas.

I know. Sometimes is quite useful, sometimes it's not

@AlekMurt You mean of their union?
$f^{-1}[A \cup B]$

@TedShifrin I have $f(x) = x^2$ from $\mathbb{R} \to [0, \infty)$ and I would like to know if i can know what it is $f^{-1}([0,1];(1,4];(4,\infty))$

11:54 PM
@Alek: So those are three separate questions!

Pick $X\in P(A\cap B')$. By definition of power set, $X\subseteq A\cap B'$. Then $X\subseteq A$ and $X\subseteq B'$

@Alek: You can't write those all in one sentence like that.

@AlekMurt What are those semicolons?

You have to ask $f^{-1}([0,1])$, then ask the next, etc.
@manooh: I'm going to write words. A subset of $A-B$ is a subset of $A$ that contains no element of $B$. Your formula is incorrect. You are removing only things that have everything in $B$, not something in $B$.

@TedShifrin what formula are you referring to?

11:57 PM
Anyhow, @Alek, what is $f^{-1}([0,1])$?

It wil be $[-1,1]$

The right-hand side of "I want to prove," @manooh. I do not believe it. Have you tried examples?
OK, @Alek, good. And $f^{-1}((1,4])$?
I'm confused. Your profile page says you've done all sorts of advanced mathematics. So why are you asking this?

What, that's what my profile says, i don't even remember wrote something like that