Mathematics - 2019-04-08

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2:12 AM

Hmmm, is there a concept of "completing a non-principal-ideal-ring?" The idea I have in mind, given a ring $R$ which is not a PIR, is adding elements to attain $\bar{R}$ which is a PIR and still has $R$ as a subring.

2:31 AM

@Rithaniel if S={1} then it's the usual maximal ideal existence theorem

2:42 AM

Ah, right, and you do need choice for that.

For a moment I thought your response was to my question about "completing a ring," and I was confused.

3 hours later…

Alessandro Codenotti

5:43 AM

@Rithaniel this is equivalent to choice

6:13 AM

Indeed I have heard that in respect to finding a maximal ideal. I just wasn't sure if it would change depending on requirements you look for on top of the ideal being maximal. Though, that's more a symptom of not really understanding the axiom of choice as well as I should. I never do notice when I'm using it until someone points it out to me.

Hi everyone. I spent a lengthy amount of time answering this post math.stackexchange.com/questions/3179063/…. OP then asked me some questions about it and then deleted the post altogether (presumably having gotten the answer they wanted). I voted to undelete. Can I request some other users with high enough reputation to look at the post and see if it is suitable for undeletion? Thanks.

1 hour later…

Tobias Kildetoft

7:16 AM

Cool. My coming employer has provided me with a 3-month membership of Pluralsight, to help me improve my skills as a developer before I start.

@TobiasKildetoft so you have a 3-month membership of Sights?

Tobias Kildetoft

yep, because that is how names work

@TobiasKildetoft hi

Tobias Kildetoft

7:31 AM

@user123 Hi

any chance i can ask you a question? it is in representation theory

Tobias Kildetoft

sure

user image

Tobias Kildetoft

If I feel like it will be either too hard or too much work, I will just leave it as an exercise for @LeakyNun

@TobiasKildetoft so i thought about $B = \{ f_{i,j} \}$

where $f_{i,j} ( x_i ) = y_j $ and $f_{i,j} (x_k) =0$

@TobiasKildetoft lol

(i assume $X=\{x_i\}$ and $Y = \{y_j\}$ )

@TobiasKildetoft it does not feel right.. i mean why would the $f$s satisfy $f(g x) = gf(x)$ ?

for $g \in G$

Tobias Kildetoft

7:36 AM

So you already know what the basis should look like in some sense, given that you are asked to find that bijection

So what does an orbit of the given action look like?

Or perhaps better, what will index the set of orbits?

$Orb(x_i ,y_j) = \{(gx_i,gy_j) \}$

@TobiasKildetoft i'm not sure i understand the question :/

Tobias Kildetoft

You have a group action. What is the usual set we use to parametrize the orbits?

as in the orbit-[blank] theorem

stabilizers?

Tobias Kildetoft

right

(this one should not be left as an exercise for @LeakyNun btw, as he will just throw Frobenius reciprocity at it and not explain how)

Alessandro Codenotti

@Rithaniel If you want to see a proof there is a simple one in a short paper by Banaschewski from 1994 that "every ring has a maximal ideal" implies AC

Tobias Kildetoft

7:49 AM

@AlessandroCodenotti Does this still hold with "maximal" replaced by "prime" (i.e. with "ring" replaced by "domain")?

Alessandro Codenotti

No, this is equivalent to the boolean prime ideal theorem

@TobiasKildetoft lol

Tobias Kildetoft

Actually, Frobenius twice makes the argument even easier

though that might require Frobenius as a bi-adjoint, which is not necessarily available I suppose

Moorrrnin

8:24 AM

@TobiasKildetoft i couldn't figure out the solution.. can you give me another hint / guidance?

Tobias Kildetoft

@user123 Did you figure out what the orbits are parametrized by?

each orbit is isomorphic to $G / stab_G (x,y)$

Tobias Kildetoft

ok, so they are parametrized by...?

by the left cosets of $G/stab_G (x,y)$ ?

Tobias Kildetoft

right

8:31 AM

$[G:stab_G(x,y) ] $ is the length of the orbit

Tobias Kildetoft

actually, no, this is not right

$|Orb(x,y)| = [G:stab_G(x,y) ] $ ain't it right?

Tobias Kildetoft

yes, that part is right

what's the difference between this and what i wrote O_o

Tobias Kildetoft

But what we really need here is to just note that the orbits are the cartesian product of the orbits of each action

the part that was not right is that this should give the parametrization (my bad)

8:34 AM

@TobiasKildetoft yes this is by the definition of the action

Tobias Kildetoft

Next we use that decomposing the set into orbits decomposes the representation into direct summands

yes $X$ is isomorphic to $\cup G/stab_G(x_i)$ where $x_i$ are representatives of each orbit

Tobias Kildetoft

Right, and this becomes a direct sum for the representation

does the basis looks like what i wrote earlier?

Tobias Kildetoft

@user123 No, because you get way too many functions that way

You only want one for each pair of orbits, not each pair of elements

8:41 AM

but doesn't the basis have cardinality of the product of basis for X and Y ?

which is $|X| |Y|$

Tobias Kildetoft

no

because it is over $F[G]$ ?

Tobias Kildetoft

right

the one you have gives all linear maps, which we don't want

if it was just F was i correct?

Tobias Kildetoft

Yes

8:45 AM

alright so $X \times Y $ is isomorphic to $\cup G/(stab_G(x_i,y_j)$

Tobias Kildetoft

Ignore $X\times Y$ for the moment

just look at each $F[X]$ and $F[Y]$ and decompose them as I said

oh. this is what you called "the representations?" ?

Tobias Kildetoft

yes, those are representations

yea you are right.. im just new to this subject :P

Tobias Kildetoft

9:02 AM

Actually, we have incorrectly determined the orbits of the product. There will be more than just the product of orbits of the two factors

Basically, each pair of orbits is stable under the action, but will generally decompose into more orbits.

9:42 AM

Where is best to refresh my skills on factorisation and simplifying equations? Any websites you recommend?

9:56 AM

@Sam brilliant.org/problems/polynomials-10

10:10 AM

Hi guys, lemme ask you something

Is 6000 in the order of 10,000 or 1000? Please answer

Cheers dude @LeakyNun

10:45 AM

Gee, I'm solving tasks in Spivaks Calculus, pretty difficult stuff

Do you think, I'll be good at one-variable calculus after solving all tasks in Spivak?

10:56 AM

Can anyone tell me abt the point at infinity of elliptic curves? The idea is not clear. Is it a set of points at infinity? Because, when we consider two points after joining which points we have a perpendicular straight line, for different points, suppose $(x_1,y_1)$ and $(x_1,-y_1)$- the line joining them will go to $\mathcal{O}$ and also line joining $(x_2,y_2)$ and $(x_2,-y_2)$ also go to $\mathcal{O}$. But both are straight line! they can't bent.

Tobias Kildetoft

@taritgoswami Are you familiar with projective geometry?

@TobiasKildetoft No :(

Tobias Kildetoft

then that is the issue. Elliptic curves live in the projective plane, which is where the point at infinity comes in

More precisely, there is a "line at infinity" in the projective plane, and exactly one point on this line lies on the curve

@TobiasKildetoft So elliptic curves we draw are not in eucledean co-ordinate system?

Tobias Kildetoft

we draw the projection to the real plane

11:01 AM

@TobiasKildetoft But, then some of the lines need to bent to reach the point at infty.

Tobias Kildetoft

I am not sure how fond I am of ever drawing elliptic curves really. It might give some intuitive idea of what is going on, but we mostly end up with the wrong idea about many things, since we usually want to really work an entirely different place

@taritgoswami There is no point at infinity when you draw the projection to the real plane

@TobiasKildetoft Ok, but can you tell me what we are projecting? Is there any material which explains all these things?

Tobias Kildetoft

@taritgoswami What we are doing is taking point in the projective real plane and projecting them onto the real plane in a "bad" way. Really we are not actually projecting (we can't). What we do instead is that we designate the line at infinity and ignore all points on this. Once we have done that, we can project the remaining point onto the real plane in a "nice" way.

This is all to do with how projective geometry works, and really learning anything of substance about elliptic curves without knowing that will be practically impossible

(though as far as I recall, the book by Washington does attempt this)

and so do Silverman/Tate it seems. They have an appendix on it, but when they introduce the point, they just gloss over it (though they do at least start by homogenizing the equation)

11:17 AM

@TobiasKildetoft Thanks, can you tell me what I need to understand Projective geom.? It seems from this video I need to understand manifolds etc. (I am an undergrad student, and don't know that adv. stuff)

Tobias Kildetoft

@taritgoswami You don't need it in that generality (and in fact, manifolds would be the wrong approach anyway for elliptic curves, where the analogue in that generality would be varieties or schemes). You just need to understand what the projective plane is

@TobiasKildetoft Ok.

2 hours later…

1:12 PM

Does $\Bbb Z[\alpha]$, the smallest ring containing $\Bbb Z$ and $\alpha\in\Bbb C$ has a special name?

That's not exactly a unique ring.

Nor is it necessarily an interesting one, e.g. $\Bbb Z[1]=\Bbb Z$

ok

Can i get some reading material on that? some link or book reference, please?

you might look at the page on the case of Z[i], i.e. the Gaussian integers:

Gaussian integer

In number theory, a Gaussian integer is a complex number whose real and imaginary parts are both integers. The Gaussian integers, with ordinary addition and multiplication of complex numbers, form an integral domain, usually written as Z[i]. This integral domain is a particular case of a commutative ring of quadratic integers. It does not have a total ordering that respects arithmetic. == Basic definitions == The Gaussian integers are the set Z [ i ] = { a + b i ∣ ...

thanks

this one also may be useful: en.wikipedia.org/wiki/Quadratic_integer

1:20 PM

ok

en.m.wikipedia.org/wiki/Kummer_ring

That looks to be the name you were looking for

WOW! That is exactly what I wanted! Thank you very much.

thank goodness for wiki "see also" links

yeah :)

1 hour later…

2:40 PM

How does $(x^2, 2x)(cosx-sinx)$ expand to $(-x^2sinx ......)$ and not $(x^2-sinx....)$

oh they are equal

1 hour later…

4:08 PM

23 hours ago, by Chebotarev Density

Of course not you lol. I was referring to you-know-who-shitposter-about-infinties

I am the Voldermort of infinity conspiracy theories

22 hours ago, by user76284

@Secret I believe that if $f : A \rightarrow A$ is injective but not surjective, then $A$ is infinite.

The problem of that is $f$ needs to map the whole $A$ but $A$ is the set we want to prove it is infinite (ok I am not really dealing with ZFC or ZF here), so there has to be a way to prove the infinitude or finitude of A without having to presuppose A be the domain in the first place

ugh, I need that book!

4:31 PM

Anyway, point is, I need to find a "local" proof of whether a set is infinite, by "local", I mean all I need is a finite subset , a procedure, and a logical proof that does the equivalent of actually running the procedure that will run forever

Such proofs do exists in some version of the problem, such as the structure of self referential sentences is the reason why we can prove by contradiction of the halting problem without actually have to go through infinite steps

5:17 PM

Suppose that a function has an infinite derivative at a point. What, if anything, can you conclude about the continuity (continuous continuation) of that function at that point?

Wouldn't that will mean it is smooth at that point? Also anything that is differentiable is continuous at that point

@Secret: I should say: the limes of the derivative of that function has an infinite derivative. So I am wondering why I cannot define a continuously continuation in the said point?

@Secret or asked differently: what does the derivative tell me about the possibility to find a continuous continuation?

If the 1st derivative is discontinuous, it means that point is a cusp, and there will be no way to find some set $P$ such that the point $p \in P$ such that $f : P \to \Bbb{R}$ be continuous

If the 2nd derivative is discontinuous, then I don't see how that will get you into trouble since it only means the curvature changes abruptly at that point. We have $x|x|$ which is continuous and any neighbourhood of the origin is also continuous, so I have no idea why it has to be smooth

books.google.com.au/…

yeah, unless I am mistaken, you only need a limit to exist at p for f to have a continuous continuation

and the continuity of the 1st derivative will guarentee that

5:39 PM

@secret: my 1st derivate looks like this $ \underset{x\rightarrow x_{0}}{\lim}t'(x)=\sqrt{\frac{2c}{t}}=\infty$. It is not continuous in 0 apparently and what you said makes sense then...

To what extent can I treat statements about linear transformations and their counterparts in matrix form ? For example when I know that ImA^2=ImA can I directly translate that to Imf^2=Imf ?

Tobias Kildetoft

@FuzzyPixelz yes

What's the rigorous explanation for that case as an example ?

Tobias Kildetoft

just a matter of writing up what the actual relation between the two things is

@Secret ... thank you for your help!!

5:45 PM

I'm afraid it's not at all trivial for me ..

Tobias Kildetoft

@FuzzyPixelz start by writing up that relation

6:05 PM

If f were the the transformation that associates to each column X the Column AX I would be fine but ..

Tobias Kildetoft

what else would $f$ be?

I pretty sure you study Maths in french, right ?

Tobias Kildetoft

nope, but I can read French math when needed

If you have any matrix A, and you consider f to be the linear map from some vector space to another, such that A would be its matrix in some basis

Then would you be able to say the same thing ?

Tobias Kildetoft

yes, $A$ being the matrix of $f$ precisely means that $f$ is given in that way (once you pick the correct basis)

6:15 PM

According to the definition I know, A's columns contain the coordinates of the images of the basis by f...

Tobias Kildetoft

yes, that is the same thing

Hmm, if you multiply A on the left by some column X that represents a vector x in our basis of choice, then you would get a column that represents f(x), is that it ?

Tobias Kildetoft

With $A$ on the left, yes

But I still find it hard to see matrix-vector multiplication and linear maps on any vector space as the same thing, even though you convinced me of their correspondence..

Wait

What am I saying ?

Nevermind, I can't believe this shadow of rigor only comes to hunts me with these things, and not when I go on hand-waving rampages

1 hour later…

7:26 PM

Hi all!

I thought about if there is a fruitful reformulation of the Töplitz conjecture en.wikipedia.org/wiki/Inscribed_square_problem in terms of group theory.

7:38 PM

One might think its: A closed curve can restrict the symmetry O(2) of the plane to at most to $D4$ in at least point.

Then the question is, does it help anything.

By the Noether theorem one might expect that some conservation property might result from this symmetry.

4 hours later…

topologicalmagician

11:31 PM

Hello

how do I prove that k is a generator for $Z_n$ iff gcd(k,n)=1

?

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