Mathematics - 2015-10-13 (page 1 of 2)

Karim Mansour

00:10

@anon

anon

?

Karim Mansour

2

Q: if $N$ normal in G, Then $\phi(N) \leq \phi(G)$, where $\phi(N)$ is the frattini subgroup of N

I was thinking somehow to use normality of N as follows Since N is normal, then $G/N \leq G$, so we can consider the natural map $pi \ :\ G \rightarrow G/N$, where $g \mapsto gN$, $ker(\pi) = N$. So It is here enough to prove that a maximal subgroup M of G will contain $ker(\pi)$, then since ...

abstract-algebra

I proved this in another way

anon

G/N<=G follows from N normal? wat?

Karim Mansour

by proving N = $phi(N)(N \cap M)$ and then proving $N = N \cap M$

but I was wondering

can I do it the way I suggested in the problem above?

that is G/N will be a group

since N is normal

anon

G/N won't be a subgroup of G

Karim Mansour

00:13

I shouldn't have typed subgroup of G

I typed that problem at 3 am yesterday

1 moment let me edit

anon

it's always a good idea to introduce trivial errors in a problem statement so you can edit it later

Karim Mansour

yeah

anon

a normal subgroup N needb't be contained in every maximal subgroup of G

Karim Mansour

yeah I was wondering if this was the case

well, the result is true I proved it other way this means that $\phi(N) \leq ker(\pi)$

since it is intersection of all maximal groups of N

I was wondering is there some kind of relation between maximal subgroups of a group and kernel of a specific homomorphism?

all we have to do is to get 1 maximal subgroup to contain the kernel as a subgroup and then we will be done

@anon I am doing a talk about frattini-subgroup so just solving couple of problems to include it in my talk

any suggestion to add about this subgroup ?

Ted Shifrin

00:33

hi @Karim @anon

Karim Mansour

01:07

Hi @TedShifrin

I am dealing with frattini subgroups

Pies

01:36

Howdy

Karim Mansour

@TedShifrin

Thomas Andrews

Did you bring enough Frattini for all of us?

Oh, wait, I'm thinking of frittata.

Karim Mansour

what completely characterizes frattini subgroup to be trivial ?

anon

well, for p-groups, it's equivalent to being cyclic

don't offhand remember things about frattini

besides it being related to elementary abelian (sub)factors in p-groups

Karim Mansour

oh yeah if its cyclic for p groups.

anon

01:45

I gave this answer over here I recall

Karim Mansour

cool thats a nice problem I will include it in my presentation

anon

it has to do with Phi(G) being minimal wrt G/Phi(G) being elementary abelian

Karim Mansour

I also saw something called prufer group

I see

anon

Since p-groups are nilpotent they have composition series, which ultimately means cyclic factors, so it's interesting how high of rank we can get in elementary abelian composition factors

what about the prufer groups?

Karim Mansour

that $\phi(G) = G$ in that case since it has no maximal subgroups

anon

01:48

they are the p-primary components of Q/Z, equivalently Z[1/p]/Z, equivalently the dual of the additive group of p-adic integers

Karim Mansour

I see

anon

yeah. prufer p's aren't special in having no maximal subgroups though

Karim Mansour

I haven't dealt with p-acdic integers before

Pies

A'right, am I the only one who thinks degrees are pointless?

Karim Mansour

it seems to me that having no maximal groups is same

the set being dense

anon

01:49

found another dim 5 subspace of H^2 characterizable by the inner product: pairs whose inner product is real. however since Sp(2) acts faithfully on H^2 and I'm trying to get a double cover of SO(5) I need a kernel +/-1 I think, so I'm going the wrong way looking at subsets of H^2.

@Pies fractions would have been nicer, just as natural and easy to understand for children. it's unfortunate.

@KarimMansour ?

Pies

Well radians are just more useful later in life

Karim Mansour

oh yeah I am wrong well I thought all countable groups will have maximal groups

Mike Miller

what's H^2

Karim Mansour

but prufer is a example of that

Hilbert space of dimension 2?

anon

H is quaternions, H^2 is a right vector space over H

Pies

01:52

Why teach degrees when they're going to be replaced?

anon

Sp(2) is 2x2 quaternionic matrices preserving the inner product on H^2

Mike Miller

@Pies in spirit, yes, but you need a degree or two to get a job ;)

Pies

.-.

Dang, that was good

anon

I haven't fully tackled qiaochu's answer but I suspect it's a translation of the clifford algebra construction (since clifford algebras are just exterior squares considered as vector spaces)

Silent

@anon, I have to show that group of rigid motion of a cube is isomorphic to $S_4$ and I can't figure out why map from one diagonal to other is injective, e.g. , let $\theta$ be rotation and $\theta(1)=2$ then $\theta (7)=8$ so $\theta \cdot\{1,7\}=\{2,8\}$ but if $\tau\ne\theta$ with $\tau(1)=8$, then also $\tau \cdot\{1,7\}=\{2,8\}$ I am working with this cube

Mike Miller

01:55

unclear if true. "exterior squares" is questionable; as a vector space it's the whole exterior algebra

anon

oops, that's right. got confused.

@Silent What's confusing you?

Mike Miller

i still don't really get the yoga of clifford algebras. they just seem to work. but eh

Silent

@anon, i cant see why the map from group of rigid motion of a cube is injective to $_4$

anon

@Silent do you see that you've failed to disprove injectivity?

you found two maps that stabilize a given diagonal. that's to be expected; there are many permutations in $S_4$ that preserve a given element of $\{1,2,3,4\}$ for instance.

waiting for a confirmation or follow-up question...

Karim Mansour

Yoga ?

haha

anon

02:06

http://mathoverflow.net/questions/64071/what-does-the-term-yoga-mean-in-mathema‌tics

Karim Mansour

oh ok that makes sense

Dan Petersen answer

Silent

02:28

@anon, but theta and tau here take diagonals to same diagonal but $\theta\ne \tau$

@anon, sorry, i was drawing image first in paper and then in paint that took time

anon

02:42

@Silent what's your point?

Silent

@anon, $\theta\ne \tau$ but they both take any diagonal to the same, eg $\theta \{4,6\}=\{1,7\}=\tau \{4,6\}$

anon

why don't you tell me what $\theta$ and $\tau$ actually are

like, the actual permutations of {1,2,3,4,5,6,7,8}

Silent

oh ok! @anon $\theta$=(1 2 3 4)(5 6 7 8), $\tau$=(1 8 3 6)(2 5 4 7)

anon

@Silent our symmetries of the cube are supposed to preserve orientation

you're looking at a bigger group of symmetries than the one intended

(rigid motions preserve orientation)

in particular, your $\tau$ does not preserve orientation

dREaM

02:59

Speaking of cubes:

$hairhairhair=cubic hair$

Silent

@anon, ok thank you!!

anon

more specifically, your $\theta$ is a clockwise rotation around the vertical axis, whereas your $\tau$ is a counterclockwise rotation around the vertical axis followed by a reflection across a horizontal plane through it

Alias User

I've been looking at this problem, but it seems poorly stated to me. If x and y are independent, don't we immediately know by definition that Pr(x|y) = Pr(x)?

dREaM

hmm

Does anyone get the joke?

$Hair times hair \times Hair= cubic Hair$

anon

$\rm hair \times hair \times hair = cubic~hair$, yes we get it

dREaM

03:03

it's a play on words

$ c \mapsto p$

anon

that's amazing

@AliasUser independence is about Pr(x,y)=Pr(x)Pr(y) isn't it?

dREaM

I have another one

Alias User

@anon - thank you!

dREaM

If two primes are twins.... incest?

@anon do you get it?

anon

I'm not sure what sex in that metaphor is supposed to be analogous to in math.

dREaM

03:06

oh damn, that doesn't translate correctly

prime in spanish is "primo". But "primo" also means cousin.

that reminds me of the time Stephen Colbert interviewed Terence Tao

anon

Tao wanted his own prime alongside Grothendieck's

dREaM

Interview With The Smartest Man In The World

►

The link starts at the funny part.

"Are cousin primes ever sexy but you're afraid to say"

anon

@MikeMiller ohlawd, to get an SO(5) action we should be acting on S^4 not S^5. If I take P^1(H) (quotient H^2-{(0,0)} by the right action of (H-{0},*) and have Sp(2) act on the left) we get S^4. wonder if it works.

we can do the same for P^1(C) to get SU(2)->SO(3), I haven't checked if that matches the Sp(1)->SO(3) I know but I have stuff to consider tomorrow.

Mike Miller

03:35

@anon yeah, in retrospect that's obvious, the Spin(6) one is the bizarre one

Karim Mansour

03:54

what math is that @anon, @MikeMiller

it seems like linear algebra stuff

and group actions

anon

geometry and lie theory

Karim Mansour

I wish I could take such a course here its not offered however

Krijn

Why does geometry always pop up?

mathamphetamines

Is this a room where I can ask a simple question without having to post it?

Krijn

Yeah, go ahead

mathamphetamines

04:04

There is a proposition in my Abstract Algebra text book on isomorphisms. It says if two groups G and G' are isomorphic, then their order tables are the same.

Krijn

Okay, and what is unclear at the moment?

mathamphetamines

I can't show that two groups have the same order table then say that they must be isomorphic correct?

Its similar to saying that just cause two groups are abelian does not imply that they are isomorphic?

Krijn

Having the same order table is much much stronger

mathamphetamines

So if two groups have the same order table does that mean that they must be isomorphic?

Krijn

Basically, the only difference between groups with the same order table is the name we give to the elements.

Yes!

Although it is often not the most efficient way to show that groups are isomorphic

anon

04:07

order table?

as in, multiplication table?

mathamphetamines

I'm working with U16 (units mod 16) and the group Z4 X Z2, so the work isn't too involved

Krijn

I hadn't even considered that you might use order table for something different, so @anon asks a very good question

mathamphetamines

Order table is a list of the possible orders of the elements then how many of those elements of a group have such orders

Krijn

Ah, no, in that case it's very different indeed, sorry for that.

mathamphetamines

For example, in U16, there is one element of order 1, three elements of order 2, and four elements of order 4. Same goes with Z4 X Z2

Krijn

04:13

I do not know if this is true or not actually. I cannot find a simple argument that would prove it or a simple counterexample to disprove it.

anon

it's not

mathamphetamines

So it is not a valid method two show that two groups are isomorphic, only a way to show that two groups are NOT isomorphic?

anon

yes. order statistics can prove nonisomorphism but not isomorphism

mathamphetamines

OK, thank you very much

anon

for instance Z/4Z x Z/4Z has the same order stats as Z/2Z x Q8, where Q8 is the quaternion group

mathamphetamines

04:18

but that doesn't show that they're isomorphic, correct?

anon

it can't show they're isomorphic, because they're not isomorphic

(one is abelian, the other isn't)

mathamphetamines

hmm, that is interesting

anon

there's really no reason to expect order statistics to determine the group up to isomorphism

Krijn

Does it hold for finite abelian groups?

I would guess so.

anon

I believe so.

Krijn

04:24

It would be nice to have some property for when this holds.

mathamphetamines

Aren't U16 and Z4 X Z2 both finite and abelian?

anon

write the number of elements of order p^r as a function of the number m(k,p) of copies of Z/p^k it has in its decomposition (as per the fundamental theorem), then solve backwards

@mathamphetamines U(16) and Z4 x Z2 are isomorphic

mathamphetamines

precisely, and @Krijn asked if the order statistics holds for finite abelian groups and you replied yes.

anon

asked if order statistics determine isomorphism type of finite abelian groups, and I replied yes

mathamphetamines

Ok, I think i misunderstood completely. Sorry for the confusion on my behalf

anon

04:29

if you can't tell that U(16) and Z4 x Z2 are isomorphic by elementary means then you shouldn't be using anything as high-powered as "ord stats imply iso type for fin ab grps."

mathamphetamines

It was given as a proposition in the section that the homework problem was assigned in, so I thought I could use it

Krijn

Ah, the beauty of mathematics.

If the group $G$ is simple, then we can determine the group by the order statistics

See also the answer to this question: math.stackexchange.com/questions/729611/…

mathamphetamines

Thanks for finding that very relevant question. I'll read it now. Looks very useful

Krijn

@mathamphetamines For your exercise I would stick to using elementary methods.

So try finding an explicit isomorphism. It creates better understanding than using heavy theorems.

mathamphetamines

sounds like a good idea

Anthony

04:44

@anon You there?

anon

yeah

Anthony

Is the image of an ideal in its field of fractions an ideal?

anon

I suppose I haven't considered fraction fields of nonunital rings before

Joe Stavitsky

anyone care to help me understand this? img.ctrlv.in/img/15/10/13/561c8b655f467.png

Anthony

What does that mean? Or rather why is that your train of thought?

anon

04:47

@Anthony well, why don't you tell me what you mean

Anthony

Yeah, sorry.

Krijn

@JoeStavitsky Write $1$ as $\frac{x^2 + 1}{x^2 + 1}$

anon

@JoeStavitsky x^2 = (x^2+1) - 1

Anthony

I'm kind of new with this field of fractions stuff- but it seems that $\frac{a}{b}\frac{i}{1}$ with $i$ in an ideal wouldn't necessarily be in the image, because $\frac{ai}{b}$ is not an element of $\frac{i'}{1}$.

BUT can't we say that $\frac{ai}{b} = \frac{aib^{-1}}{1}$?

anon

@Anthony start at the beginning and tell me what the setup to your question is

"Let R be a ring..."

you cannot expect me to read your mind from possibly thousands of miles away

Joe Stavitsky

04:49

ahhh so ty

anon

your latest comment uses i to refer to "an ideal" (who uses lowercase i for an ideal?) but also apparently refers to an element, then writes i' without saying what that is, et cetera

Anthony

Sorry, I'm writing something now.

Let R be a ring, and I be some ideal of R. Consider the image of I under the map $f: i \in I\to \frac{i}{1}$. Is $f(I)$ an ideal in the field of fractions $Frac(R)$?

anon

no

Anthony

Okay, what if $R$ is commutative?

anon

I thought we were assuming R was commutative, but whatever.

Anthony

04:52

Oh, okay. I just don't know what the hell is going on with algebra.

anon

we take fraction fields of integral domains (commutative rings with no zero divisors)

we can also take localizations more generally

Anthony

Okay, let me back up my tree of problems.

I was looking at Dedekind rings, and the problem I'm working on says given an ideal, consider its inverse.

But I wasn't sure why an arbitrary ideal would have an inverse.

anon

assuming R is a domain, then Frac(R) is a field. the only ideals of fields are (0) and the whole field. so I remains an ideal in Frac(R) iff I=R and R is already a field.

Anthony

I see.

Duh.

anon

@Anthony would it be talking about its inverse as a fractional ideal?

Anthony

04:55

"Given an ideal $\mathfrak{a}$ let $\mathfrak{b}$ be the fractional ideal such that ab = $\mathfrak{o}$, where $\mathfrak{o}$ is a Dedekind ring."

anon

right

Anthony

Why do we have the existence of such a b?

anon

that takes some work to prove

Anthony

Oh. I didn't see it in the text, but it's probably there then.

anon

but for example with Z, the inverse of nZ would be (1/n)Z inside Q

consult almost any text on algebraic number theory for discussion of fractional ideals

Anthony

04:57

You're pretty knowledgeable for an undergrad.

Or I'm pretty dumb.

Or some combination of both.

Thanks for the help.

anon

I was out of school for a few years but still taught myself math.

Anthony

Oh, I see.

Anyway at the moment I'm looking at Lang's Algebra.

But so is the existence of such a $\mathfrak{b}$ something unique to Dedekind Rings?

anon

yes

Anthony

I see- yeah it seems like he only mentions Dedekind Rings in passing...

anon

a domain is dedekind iff every ideal is invertible as a fractional ideal

Anthony

05:03

I thought it was iff every nonzero fractional ideal of R is invertible.

Are those two statements equivalent?

anon

probably. or I could be misremembering. (hey, if you already know the iff, then why are you asking me?)

Anthony

Wait, sorry. I'm asking because I'm terribly confused. That said every fractional ideal is invertible.

But I was wondering why every ideal was invertible, which is what Lang seemed to be implying.

And I couldn't find that anywhere.

anon

if fractional ideals are invertible then certainly normal ideals are

Anthony

Oh, are normal ideals fractional ideals?

anon

check the definitions

Anthony

05:07

On it.

Thanks.

Jessy Cat

heya

Extremely overworked grad student.

How do I stay sane????

crocket

Graduate work is insane.

Anthony

05:36

@anon Wait are you still there?

Jessy Cat

08:32

0

Q: Outer measure of empty set is 0 (using definition of outer measure)

I am trying to show rigorously that $m^{*}(\emptyset)$ is zero. I know intuitively why it makes sense, but intuitively isn't going to cut it - I need to show it rigorously where $m^{*}(A) = \inf \left\{ \sum_{k=1}^{n}l(I_{k}) | I_{1},\cdots,I_{n}\,\text{open intervals,}\,A \subseteq \cup_{k=1}^{n...

real-analysis measure-theory lebesgue-measure

Hey

You must be Steffan ;)

Hi @DanielFischer

Daniel Fischer

Hi.

Balarka Sen

09:30

@anon er? i have not been following the discussion, but it's not true that order stats determine isom type of finite abgrps. Z_4 and Z_2 x Z_2.

but i guess you had order stat of the torsion subgrps in mind when you said that?

Joel

10:09

I have a homework in a first course in Mathematical Logic and I wonder if there's any difference between $\neg\exists x\varphi$ and $\neg(\exists x\varphi)$ because if this is the case, which I think it is, I think it's pretty weird to have the last mentioned as a homework since the first one is an example in my Logic book... It's an easy one to solve indeed if they are the same, I solved it as they did in the book but still...

Btw my goal was to derive the formula (predicate logic) forgot to mention it ;)

*Btw my goal was to derive a formula using that as an ASSUMPTION.,,

Manolis Lyviakis

12:22

0

Q: Find the self-interection.Differential Geometry

Show that the Cayley sextic $$γ(t) = \bigl(\cos^3 (t)\cos (3t), \cos^3 (t)\sin (3t)\bigr),\quad t \in \mathbb R,$$ is a closed curve which has exactly one self-intersection. What is its period? I can see $2π$ is its period.But for the self-intersection have to solve $γ(a)=γ(b)=p$ ?

calculus multivariable-calculus differential-geometry

help?

crocket

12:43

What do I need to learn to do logic programming in miniKanren?

Tobias Shuxue Laoshi

Hey people! Question: is -\frac{\zeta'(s)}{\zeta(s)}-\frac{1}{s-1} entire? It ought to be, since it doesn't have any poles. Amirite?

Daniel Fischer

@TobiasShuxueLaoshi No, it has poles at the zeros of $\zeta$.

Tobias Shuxue Laoshi

Check it again :)

\zeta' = g'/(s-1)-g/(s-1)^2 for g holo at 1.

zeta'/zeta = ((s-1)/g)( g'/(s-1)-g/(s-1)^2)

((s-1)/g)( g'/(s-1)-g/(s-1)^2)=g'/g-1/(s-1)

Remember me

You can use Latex @TobiasShuxueLaoshi

Tobias Shuxue Laoshi

Hence, -\frac{\zeta'(s)}{\zeta(s)}-\frac{1}{s-1} = -\frac{g'}{g}+\frac{1}{s-1}-\frac{1}{s-1}

Daniel Fischer

12:54

@TobiasShuxueLaoshi But at a zero of $\zeta$, you have $\zeta(s) = (s - z_0)^k\cdot h(s)$ with $h$ holomorphic and nonzero in a neighbourhood of $z_0$. Then $$\frac{\zeta'(s)}{\zeta(s)} = \frac{k}{s-z_0} + \frac{h'(s)}{h(s)}$$ has a simple pole with residue $k$ (the multiplicity of the zero) at $z_0$.

Tobias Shuxue Laoshi

But hey hey, look at what I wrote. We rid ourselves of it.

-\frac{g'}{g}+\frac{1}{s-1}-\frac{1}{s-1}=-\frac{g'}{g}

Which is holo at $s=1$.

And hence, since the only pole of riemann zeta is the one at $s=1$. We ought to have that $\frac{g'}{g}$ is holo everywhere.

Daniel Fischer

@TobiasShuxueLaoshi At $s = 1$, but it's not entire.

Tobias Shuxue Laoshi

Why?

Daniel Fischer

Because the zeros of $\zeta$ give rise to poles of $\frac{\zeta'}{\zeta}$, see above.

Tobias Shuxue Laoshi

Sure, ok, the trivial zeros. And whatnot, meh.

Thanks, that was what I needed.

But it is holo to the right of one :) Right?

Daniel Fischer

13:00

@TobiasShuxueLaoshi Yes. And probably for $\operatorname{Re} s > \frac{1}{2}$.

Tobias Shuxue Laoshi

Hehe

crocket

Can anyone answer math.stackexchange.com/questions/1478121/… ?

Tobias Shuxue Laoshi

@Daniel Fischer: Hey Daniel, $\zeta=g/(s-1)$ with $g$ entire, though?

Daniel Fischer

@TobiasShuxueLaoshi Yes, $g(s) = (s-1)\zeta(s)$ is entire (after removing the removable singularity).

Tobias Shuxue Laoshi

@DanielFischer: Well that's kinda nice at least :)

Tobias Shuxue Laoshi

13:26

Daniel, as you seem to be knowledgeable on analytic number theory, I'd like to ask; which is your favorite near but not quite graduate-level textbook on analytic number theory. I'm using Apostol's and the proofs aren't so good. Or maybe I just haven't learned to appreciate them.

But there are weird things in it. Like, he shows something super-basic from complex analysis right at the end of his proof of PNT. It kinda messes with the pace. I think.

Remember me

Hello@Balarka

Balarka Sen

hi

@Remember Have you proved that exercises you were asking me about? That any two maps $f, g : Y \to CX$ are homotopic?

Remember me

I just saw your comment. I might try it today night.

Balarka Sen

By the way, $X$ is connected, right? Otherwise it's not necessarily true.

Nevermind. It's of course always true. I was thinking of the suspension.

Secret

Balarka Sen

13:38

@Remember The thing you'll need here is the following fact : $CX$ is contractible for any topological space $X$. Do you know how to prove this?

Secret

Are there any useful cases where we will come across expression of the form:
$$g(x)=af(x)+bf(f(x))+cf(f(f(x)))+\cdots$$?

(Something kinda like a iterative map version of a power series)

Remember me

Well if it is contractible then I have to show there is a homotopy between the identity map and the constant map .

Balarka Sen

Mhm.

Daniel Fischer

@TobiasShuxueLaoshi Not really, picked up a couple of things in passing. Don't know the literature at all.

Remember me

This is a question in Armstrong and I don't think so it is given that X is connected

Balarka Sen

13:41

Yes, apologies, it's true for all spaces.

Tobias Shuxue Laoshi

@DanielFischer: Oh, I see. Well, I'll see what I think about all the others ones when I start going through them. Apostol has been good as a start though.

Remember me

Well now coming back to the center question:
A loop $f$ in the center of $\pi_{1}(X,x_0)$would mean that $[f][g]=[g][f], \forall [g]\in \pi_{1}(X)$ . Now what I did till now that I have considered another at $x_0$ and in $\pi_{1}(X)$ and the trying to find a homotopy between $[f][g]$ and $[g][f]$ and also between $[g][f]$ and [f][g]$ . The thing I am thinking is that if I can somehow draw a picture of these two maps then I can find a suitable homotopy and I am done with it@Balarka

N3buchadnezzar

13:58

Heya

A student i am TA for was asked to find $n$ such that the maclaurin expansion of $\sin x$ had a error of less than 0.001 at $x = \pi$.

This basically boils down to finding $n$ such that
$$ \frac{ \pi^{2n - 1} } { (2n-1)! } \leq \frac{1}{1000} $$

Is there some way to estimate this without using a calculator?

Daniel Fischer

@N3buchadnezzar You can use $\pi \approx \sqrt{10}$ to simplify the computations when you're doing them without calculator.

Secret

Why it looks as if there are 3 humps here

Also if I start my initial condition at osme earlier point on the lorenz attractor, do I expect it to trace out a completely different lorenz attractor?

That is, if I start from B instead of A, since they both lie on the same trajectory, will I expect the final lorenz attractor with intial conditons A and B to be very different form each other?

@Gato @DanielFischer Which regular in this chat is good at chaos theory and dynamical systems?

Daniel Fischer

14:22

No idea, don't remember that coming up before.

Galois in the Field

What does $F(\alpha)$ look like, where $F$ is a field, and $\alpha$ is an adjoined root?

Krijn

14:46

What is the degree of $\alpha$?

@GaloisintheField

Galois in the Field

What does degree mean in this context sorry?

Krijn

The degree of the extension $[F(\alpha) : F]$

Or equivalently the degree of the minimal polynomial $f$ such that $f(\alpha) = 0$

Galois in the Field

Basically my question comes down to, does $F(\alpha)=a_0+a_1\alpha+a_2\alpha^2+a_3\alpha^3+\cdots$, because $\Bbb Q(\sqrt 2) = \{a+b\sqrt{2}:a,b\in \Bbb Q\}$ but that is pretty much the reduction of:

$$a_0+a_1\sqrt{2}+2a_2+2a_3\sqrt{2}+\cdots$$

Krijn

Basically it does, but it is the reduction exactly because $\alpha^n = \sum c_i\alpha^i$

Which we get from the minimal polynomial

(for the right $c_i$ of course with $i < n$)

Galois in the Field

Because $\alpha^n = \sum c_i\alpha^i$?

Krijn

14:53

We know that we have some minimal polynomial $f$ such that $f(\alpha) = 0$

Galois in the Field

Sure

Krijn

So there is some $\sum c_i X^i$ such that $\sum c_i \alpha^i = 0$

Where the index of the sum must be finite

Galois in the Field

Oh I see

Krijn

And the degree $n$ of the minimal polynomial of $\alpha$ is equal to the degree $n$ of the extension

Galois in the Field

is that always true?

Krijn

14:55

Which means that we would get a base of order $n$, which is exactly $\{ 1, \alpha, \alpha^2, \ldots, \alpha^{n-1} \} $

It is if it is an extenstion of adjoining only one root

Think about this for a second, I´ll be back in a minute

Galois in the Field

Okay I shall, I have another question when you get back

(if you are willing of course)

Mary Star

I have to find an example that the reparametrization of a closed curve is not necessarily closed.

Do you have an idea how to find such an example? @robjohn @Huy

Krijn

15:10

@Galo

@GaloisintheField Back, but gone in like 5 mins

I'm out, but if you @ me I'll respond in like an hour or so

Mary Star

15:29

Do we have to use trigonometric functions? @robjohn @Huy @DanielFischer @anon

Do you maybe have an idea about my question about? @TedShifrin

robjohn

@MaryStar what definition of a curve are you using? is it a continuous map from $[0,1]$ to $\mathbb{R}^2$?

Mary Star

In my book it is $\gamma : \mathbb{R} \rightarrow \mathbb{R}^n$ or a subset of $\mathbb{R}$, i.e., $\gamma : [a,b] \rightarrow \mathbb{R}^n$. @robjohn

And we assume that the curves are smooth. @robjohn

robjohn

@MaryStar A closed curve is usually defined as a smooth function $\gamma:[0,1]\to\mathbb{R}^n$ where $\gamma(0)=\gamma(1)$.

Mary Star

In my book there is the following definition:

Let $\textbf{$\gamma$}: \mathbb{R} \rightarrow \mathbb{R}^n$ be a smooth curve and let $T \in \mathbb{R}$. We say that $\textbf{$\gamma$}$ is $T$-periodic if $$\textbf{$\gamma$}(t+T)=\textbf{$\gamma$}(t) \text{ for all } t \in \mathbb{R}.$$
If $\textbf{$\gamma$}$ is not constant and is $T$-periodic for some $T\neq 0$, then $\textbf{$\gamma$}$ is said to be closed.

@robjohn

Galois in the Field

@Krijn I have to go aswell, but I'll read whatever response you give me and appreciate it greatly.

So if I have a degree four irreducible polynomial $f(x)$ over a field $F$ and I adjoin one of its roots $F(\alpha)$ and this is the splitting field of $f(x)$.

Then $F(\alpha)$ is a $F$-vectorspace with basis $\{1,\alpha\}$. Then any $F$-vectorspace automorphisms will take $\alpha$ to the second root of $\alpha$'s minimal polynomial right?

Just trying to understand the alternative root part of this answer:

http://math.stackexchange.com/questions/1472951/what-is-a-k-automorphism-of-l

(to one of my questions)

robjohn

15:43

@MaryStar Unless you change the domain of the function so that the range does not cover the whole of the curve, I don't see how one could reparametrize a closed curve to a non-closed curve.

Mary Star

And if we change the domain of the function, how can the reparametrization of a closed curve be non-closed? @robjohn

Secret

Can we have a parametrisation that is discontinuous?

e.g. if the curve a circle, can we do this (so that point where the curve =1 is there just to make it open?)

Mary Star

Can we maybe take te curve $\gamma (t)=(\cos t, \sin t)$ and the reparametrization the $\gamma (\ln t)=(\cos \ln t, \sin \ln t)$ ? @robjohn

Secret

So if $\theta \in [0, 2\pi]$ then for $\theta \in [0,a]$ where $a<2\pi$, we have basically parametrised the original curve, and when $\theta=2\pi$, that point is deliberately offset so that under this parametrisation the curve becomes open at $\theta=0$? (since $\theta$ has to run from 0 to $2\pi$ to define the curve)

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