Mathematics - 2018-08-23 (page 1 of 2)

Semiclassical

12:00 AM

"second verse, same as the first!"

manooooh

I don't want "because the sum of two squares is always positive"

Semiclassical

it's not.

it's never negative != is always positive

manooooh

@Semiclassical ahhh yes, exclude the case $x=y=0$. Sorry

Leaky Nun

finally

Semiclassical

So you want to prove that $x^2+y^2=0$ iff $x=y=0$.

Leaky Nun

12:02 AM

@manooooh if you want to prove it using axioms, note that this is true more generally in ordered fields

Semiclassical

(the 'only if' direction is of course the only nontrivial one)

manooooh

No, I want to prove that $\forall x,y\in\mathbb R^2\setminus\{0\}$ then $x^2+y^2>0$

Semiclassical

ah, was reading your original version first

Leaky Nun

Ordered field

In mathematics, an ordered field is a field together with a total ordering of its elements that is compatible with the field operations. Historically, the axiomatization of an ordered field was abstracted gradually from the real numbers, by mathematicians including David Hilbert, Otto Hölder and Hans Hahn. This grew eventually into the Artin–Schreier theory of ordered fields and formally real fields. An ordered field necessarily has characteristic 0 since the elements 0 < 1 < 1 + 1 < 1 + 1 + 1 < ... necessarily are all distinct. Thus, an ordered field necessarily contains an infinite number of...

manooooh

@LeakyNun thanks! I didn't see fields, maybe using some properties of the real numbers?

Leaky Nun

12:04 AM

I already told you

the relevant properties are called the ordered field axioms

manooooh

@LeakyNun ok, I will review it and if I have doubts I will post a question

geocalc33

12:30 AM

remind me, why in complex analysis do we use the formula $e^{nlogx}=x^n$

mick

-2

Q: How is the g-factor $ 2.002319 $ computed?

I understand a lot of math but not so much physics. So I read about the feynman diagrams and QED and I started to wonder : How is the g-factor $ 2.002319 $ computed exactly ?? Do we use matrices , differential equations , ... ?

mercio

because that can be used for $n$ that is not an integer

mick

Help ?

Daminark

Sniped

Semiclassical

Feynman diagrams basically are a convenient tool for enumerating multiple integrals

mick

12:32 AM

1

Q: Prime twins and $ f(z) = \sum_p \frac{\ln(p) \space \ln(p+2)}{p^{z/2} \space (p+2)^{z/2}} $

Let $p$ be a prime such that $p+2$ is Also a prime. Define $$ f(z) = \sum_p \frac{\ln(p) \space \ln(p+2)}{p^{z/2} \space (p+2)^{z/2}} $$ For $z \neq 1 $ and $re(z) > 1$ this $f(z)$ always converges. If there are infinitely many prime twins then clearly $f(1)$ diverges. If there are only fin...

number-theory analytic-number-theory conjectures zeta-functions prime-twins

geocalc33

so it's used to define $x^n$ for values of $n$ that are non integers. interesting

Semiclassical

so computing the g-factor amounts to writing out a lot of multiple integrals in the right way and computing them

being more detailed than that requires a lot more know-how than I actually possess, though

(also, this applies in the framework of perturbative QED. nonperturbative QFT is harder)

loch

it's the ring of regular functions on the variety yes! Let's say you're considering the variety in $\mathbb{C}^n$ defined by equations $f_1,\ldots,f_m = 0$, then if you think of the polynomial (regular) functions defined on $\mathbb{C}^n$ they're given by $\mathbb{C}[x_1,\ldots,x_n]$. Now if you restrict all such functions to the locus defined by $f_i=0$, then e.g. $g$ and $g+f_i$ will restrict to the same function on the zero locus of $\{f_i=0\}$. So the ring of regular functions is given by $\C[x_1,\ldots,x_n]/(f_1,\ldots,f_m)$.

geocalc33

so for example, the equation of a circle could be defined as $e^{nlogx}+e^{nlogy}=1$, which simplifies to $x^n+y^n=1$ right?

oh wait that's a pointless definition

it's like a pencil without a point

pointless

@geocalc

I want to ping myself I'm taking this up on META

Adam

1:21 AM

They don't need the "hot network questions" panel on the RHS most of the questions I see there are definitely written by people under the influence of drugs and totally not related to mathematics!!!!!!!!!!!!!!!!!!!

geocalc33

@Adam is there any point in defining the equation of a circle as $e^{nlogx}+e^{nlogy}=1$

well its actually superellipses with $n=2$ being a circle

my name is Hertzcog Werner pronounced "Verner"

@XanderHenderson is there any point in defining the equation of a circle as $e^{nlogx}+e^{nlogy}=1$

Xander Henderson

Well, that won't generally be a circle, @geocalc33

geocalc33

drangus fandangus...

but for $n=2$

Xander Henderson

well, yes, that would be a circle for $n=2$

geocalc33

but why would anyone define it like that?

Xander Henderson

1:32 AM

define what?

geocalc33

like you could just write $x^s+y^s=1$

instead of using the exponentials

Xander Henderson

A circle is typically defined to be the set of points that are a fixed distance from a distinguished point called the center.

geocalc33

$e^{nlogx}+e^{nlogy}=1$

Xander Henderson

It can be shown that that the set of solutions to the equation $(x-h)^2 + (y-k)^2 = r^2$ is a circle.

geocalc33

i know that

Xander Henderson

1:34 AM

Okay, but that isn't the definition of a circle (usually) any more than the equation $y=mx+b$ is the definition of a line.

geocalc33

in complex analysis you define $x^n$ as $e^{nlnx}$

Xander Henderson

And just as we can write $Ax + By + C = 0$ or $y-y_0 = m(x-x_0)$, you could also write $\mathrm{e}^{2\log(x)} + \mathrm{e}^{2\log(y)} = 1$ if you wanted.

@geocalc33 Do I?

geocalc33

yes that's a common definition

i read about it

Xander Henderson

Do you understand why one might want to do that?

And do you also understand how that definition works if $n$ is an integer?

geocalc33

to extend $n$ to non integer values?

yeah you can differentiate $x^n$ and you get $nx^{n-1}$

Xander Henderson

1:38 AM

I mean, another possible definition of complex exponentiation is that $x \mapsto x^s$ is the unique analytic continuation of the the usual real valued exponential function on $(0,\infty)$, continued to $\mathbb{C}$ (minus, perhaps, a branch cut).

Where we can define the real exponential as the unique continuous extension of the rational exponential, and the rational exponential can be defined in terms of integer roots and powers.

geocalc33

okay so what does all that mean in terms of the equation $e^{2logx}+e^{2logy}=1$

Xander Henderson

(where by "exponential," I really mean "power functions"; I should be more careful in my use of language)

@geocalc33 I don't understand what you are trying to get at with that equation.

The set of real solutions to that equation is a circle.

geocalc33

what about complex

Xander Henderson

And so?

It is more complex in complex land, since the exponential possesses some periodicity.

A circle in $\mathbb{C}$ is typically written like $|z| = 1$ (or, more generally, $|z-\alpha| = r^2$).

geocalc33

i don't know it's just a really interesting equation

Xander Henderson

1:43 AM

@geocalc33 WHY is it interesting?

It is the usual equation for a circle, complexified, and written in a kind of funny manner.

geocalc33

because the domain and codomain vanish outside the unit square

for non negative $n$

so it basically gets rid of the other part of the circle

and just gives you the part that passes through the unit square

Adam

@geocalc33 are you referring the question I just posted?

geocalc33

which one is that?

Xander Henderson

You haven't defined any functions. You have given an equation. What domains and codomians are you talking about???

Adam

No it's just that you were talking about the unit circle and I did post a question that involves the typical identities that ends up having something to do with that

and you asked me specifically so I assumed it was for that post since it's probably my least effort one in a while so there is likely to be something stupid in ir

mainly it's because maple wont share their inbuilt code for how they compute floating point approximations, so every time they have inbuilt code that contradicts other inbuilt code, I'm just going to somehow find a way to complain about it as stack exchange question until they let me debug inbuilt code, so I no longer have a blind spot

its very frustrating to 85 percent of the time be able to trust that the approximations are accurate, and then all of a sudden have it tell you what you have done is false, when you are right, or vice versa

I got a free large iced coffee frappe from maccas drive thru this morning.

geocalc33

2:04 AM

cool

Adam

it was a good day overall

geocalc33

nice mine was too

Xander Henderson

I did not think to coldbrew last night, and ended up getting terrible, terrible coffee on campus at Charbucks this morning. That sucked. :\

I need to get a second coffee maker for my second office so that I don't have to drink terrible coffee.

geocalc33

@XanderHenderson I was just trying to understand that equation

it seems it's not really that interesting

Xander Henderson

@geocalc33 I get that you are trying to understand the equation. I just don't understand what aspect of it you aren't understanding. It is just the equation of a circle, complexified, and written funny.

geocalc33

2:10 AM

yeah I understand that part I was just trying to find somehting interesting about it that wasn't there

Xander Henderson

Just like $\alpha w + \beta z = \gamma$ is an equation for a line, complexified, and written funny.

geocalc33

that's what I was plotting

Silent

Is it true that set of all condensation points in a separable metric space $X$ is either empty or uncountable?

geocalc33

and I was thinking that the red line could be the identity element and on either side you have your elements and inverse elements

Xander Henderson

@geocalc33 of course that is what desmos plotted---desmos doesn't know about complex numbers

geocalc33

2:14 AM

anyway..

okay but now I'm just talking about reals

Xander Henderson

wolframalpha.com/input/…

geocalc33

could it be a group though? with identity element as $(1-x)+y=1$

user10478

How can you find the summation of k = 1 to infinity of 3/(4^k) via telescoping?

Xander Henderson

@geocalc33 What is "it"?

Adam

anyway it shouldn't have a problem verifying $9\,\pi \,{{\rm e}^{3}}-180\,\pi =9\,\pi \, \left( {{\rm e}^{3}}-2
\right) -162\,\pi $

Xander Henderson

2:19 AM

Could what be a group?

What is the underlying set, and what is the group operation.

Adam

Like the software does simplifications like that all the time why there would it suddenly need me change the settings

geocalc33

@XanderHenderson the set of elements consist of: $e^{slnx}+e^{slny}=1$. The set of inverse elements are those reflected about the identity element which is $e^{lnx}+e^{lny}=1$. I'm not sure what the group operation is though...hmm

addition? multiplication? composition?

Xander Henderson

@geocalc33 A group is defined by its elements and an operation; you don't need to specify inverse elements.

geocalc33

okay

Xander Henderson

Once you define the elements and the operation, you can determine the inverse elements from the definition.

geocalc33

2:25 AM

I guess I'll go with addition

when the operation is applied to any two group elements it should give another element in the group

hooray

it works

do you concur @XanderHenderson

Drew Brady

3:24 AM

Anyone feeling like helping me think about a measure theory problem? :)

Luc

4:52 AM

I think I've reasonably described my question, but I find it hard to think of good tags. Looking through common tags, I'm having a "mhm, I know some of these words..." moment (the math education I had, which isn't much, wasn't in English). Could someone look over the tags of my question to see if there are any missing and whether the ones I chose are correct? math.stackexchange.com/questions/2891706

dude3221

5:06 AM

hello

@Luc

sorry I am not qualified to help you properly

Alessandro Codenotti

7:19 AM

@DrewBrady what's the problem?

Rudi_Birnbaum

8:00 AM

Good morning all!

When I have two groups $G,H$ and I form new group by direct product formation, I do not see immediately that $|G\times H| = |G||H|$. Because its not immediately clear to me that $(g_1,h_1)=(g_2,h_2) \Rightarrow g_1=g_2 \land h_1=h_2$ ($\Leftarrow$ is clear). The new group potentially could have a different "identification operation", no?

Tobias Kildetoft

8:20 AM

@Rudi_Birnbaum By definition, just as a set, we have that those orders agree

i.e. that property of when two pair are identical are part of the definition of the direct product as a set

Shobhit

What are the equivalent definitions for a set to be nowhere dense in R, all i know is the interior of the closure of the set is null

Abcd

Differential equation of all non horizontal lines in a plane?

Attempt:

$y = mx + c$, $m \ne 0$

$y' = m$

$y''= 0 $

But answer given is:

$\dfrac{d^2x}{dy^2} = 0 $

Why am I wrong?

Alessandro Codenotti

@Shobhit not dense in any open set

But I don't know how many characterisations are there or what are you looking for

Shobhit

8:35 AM

So, if $A$ is nowhere dense in $R$ then there exists an open interval $O$ such that $O \cap A = \phi$ @AlessandroCodenotti

Alessandro Codenotti

No, but $O\cap A$ is never dense in $O$

Drew Brady

8:57 AM

@AlessandroCodenotti it was math.stackexchange.com/questions/2891667/…

But I have since figured it out

Leaky Nun

9:14 AM

@Shobhit this is right @AlessandroCodenotti

but the converse is not right

I believe a characterization is that every open set O contains an open set U such that U intersect A is empty

Shobhit

9:27 AM

thanks @LeakyNun

@AlessandroCodenotti

Alessandro Codenotti

9:51 AM

@LeakyNun My bad, I wasn't clear, I meant that this is not equivalent to being nowhere dense and is not equivalent to not being dense in any open set. But it's true of course

Abcd

10:24 AM

@LeakyNun Are you there

Leaky Nun

?

Abcd

2 hours ago, by Abcd

Why am I wrong?

Leaky Nun

@Abcd you aren't wrong

Abcd

2 hours ago, by Abcd

But answer given is:

2 hours ago, by Abcd

$\dfrac{d^2x}{dy^2} = 0 $

Leaky Nun

just express x as subject of y

ok in another sense, the given answer captures the "non-horizontal" part which your answer does not

Abcd

10:41 AM

@LeakyNun ?

Abcd

11:22 AM

@LeakyNun Can you please elaborate?

Can any parabola have both x^2 and y^2 terms?

Leaky Nun

u^2 = 4v is a parabola

substitute u=x+y and v=x-y

Abcd

@LeakyNun Hmm, what about previous question?

Leaky Nun

that's a rather ridiculous question

Abcd

:/

NaCl

11:40 AM

Hi guys

geocalc33

hi

Secret

12:05 PM

@Abcd if it is f(x,y), we call that a parabloid, otherwise if it is y^2=f(x), it's something else entirely

$ax^2 + bxy + cy^2 = d$ are conics

Abcd

@Secret Didnt you see leaky nun's example?

desmos.com/calculator/indcuouaod

@Secret Is that not a parabola^?

@LeakyNun can you help me figure out mistake in my solution of a question?

Leaky Nun

?

Secret

@Abcd O, I was thinking about squared terms with negative coefficients, yes that one is a parabola rotated

Abcd

@LeakyNun Order and degree of differential equation whose solution is: $(x-h)^2 + (y-k)^ = a^2$

Solution:

$x- h + yy' - ky' = 0 $

$\implies 1+ yy'' + (y')^2 - ky'' = 0 $

So order=2

degree = 1

But answer is:

Order 2

Degree 2

Leaky Nun

@Secret ^

Abcd

12:18 PM

@LeakyNun Whats my error?

@LeakyNun Why did you ping him/her?

Leaky Nun

because I have no idea

Secret

Why is the equation being differentiated twice, are they asking for a 2nd order equation?

Abcd

Find: Order and degree of differential equation whose solution is ...

Secret

if you differentiated that once, you get a 1st order degree 1 nonlinear ODE
differentiated that again, you get a 2nd order degree 2 nonlinear ODE

Order is the largest derivative, while degree is the largest exponent

Abcd

@Secret Degree is largest exponent of highest order derivative which in this case is y'' whose exponent is 1

Never mind. Ill ask on main.

Secret

12:31 PM

0

Q: Question about the degree of differential equation

Is that the correct answer in the picture? According to the definition, The degree of a differential equation is the power of the highest order derivative in the equation, so my question is why the result is not 1 (because the first term is the highest order derivative) but 3?

calculus

and it does not help that degree is only well defined if the y form polynomials

Rudi_Birnbaum

@TobiasKildetoft OK, thanks! (It was not explicitly mentioned in my textbook, afaiu (Hungerford))

geocalc33

12:52 PM

-2

Q: Group under addition?

Is this a valid group? I tried going through all the conditions. $G$ is the set of equations for $x,y \in \Bbb R(0,1); $ $G= \{ e^{s{\rm ln}(x)}+e^{s{\rm ln(1-y)}}=1; s\in \Bbb R \}$ with addition as the group operation. So for example adding two elements of the group together: $ e^{2{\rm ln...

abstract-algebra group-theory

@Rudi_Birnbaum do you understand that question

Tobias Kildetoft

@geocalc33 That question it complete nonsense as far as I can tell.

geocalc33

no it isn't

.

Tobias Kildetoft

@geocalc33 Ahh, you are the same person as the OP?

geocalc33

no

Tobias Kildetoft

Then why are you claiming it is not complete nonsense?

Rudi_Birnbaum

12:55 PM

$e^{x ln(y)}$= y^x, to begin with. Then $\Bbb R(0,1)$ is odd $(0,1)$ might do as well ... I didn't read further yet

geocalc33

okay fine i am the same as the op

for some reason I'm still geocalc in the chat though

user 2646

BUSTED!!! :P

Rudi_Birnbaum

Is there an identity element and an inverse, for each?

geocalc33

there is an inverse element for each element. To get each inverse element just reflect the element over the identity element

Rudi_Birnbaum

for $y=1$ and $s=-1$ you get division by zero ...

geocalc33

1:01 PM

y is not defined for 1

y does not include 1

$y \in (0,1) $

Rudi_Birnbaum

@geocalc33: How should $x^{s_1} + (1-y)^{s_1} + x^{s_2} + (1-y)^{s_2}$ be identical to $x^{s_3} + (1-y)^{s_3}$ for any $s_3$? Freshmans dream?

Freshman's dream

The freshman's dream is a name sometimes given to the erroneous equation (x + y)n = xn + yn, where n is a real number (usually a positive integer greater than 1). Beginning students commonly make this error in computing the power of a sum of real numbers. When n = 2, it is easy to see why this is incorrect: (x + y)2 can be correctly computed as x2 + 2xy + y2 using distributivity (commonly known as the FOIL method). For larger positive integer values of n, the correct result is given by the binomial theorem. The name "freshman's dream" also sometimes refers to the theorem that says that for a prime...

user 2646

Rudi_Birnbaum

@user2646 yeah!

geocalc33

well...

let me just put my thinking cap on

maybe you just BUSTED my group

anyway for $s \in \Bbb R(1,\infty)$

Rudi_Birnbaum

@geocalc33 No worry I didn't bust it, it has never been one ...

geocalc33

1:11 PM

yeah it is

Secret

in The h Bar, 3 mins ago, by danielunderwood

Is there a sort of abstract version of a tensor that you could call type (a, b, ..., k) with k types of indices? I can't see how it would come out of the geometric definition of a tensor, but I wondered if there was a way to create such a thing.

Trivariant tensor ftw

Rudi_Birnbaum

@geocalc33 You again don't understand you own definition: it should hold for all x,y of your domain, while the $s$ are the "indices" for the elements thus fixed for each element ...

Secret

Pairs well with Triglavian, Triangulum, Tripodium

geocalc33

@Rudi_Birnbaum how does it not hold?

Rudi_Birnbaum

its not closed

if you disagree, tell me what is $s_3$?

(as a function of $s_1$ and $s_2$)

geocalc33

1:18 PM

okay one second

Rudi_Birnbaum

relax and take 2

geocalc33

$e^{2{\rm ln}(x)}+e^{2{\rm ln(1-y)}}+e^{3{\rm ln}(x)}+e^{3{\rm ln(1-y)}}=2$

Rudi_Birnbaum

(except $s_1=s_2=0$ that should work, then your group has one element ...)

@geocalc33 What???

geocalc33

You add the equations together!!!!

lol

Rudi_Birnbaum

@geocalc33 That is not $e^{s_3 ln(x)} + e^{s_3 ln(1-y)} = 1$, but that is what should result, to have at least something looking a bit like closedness ...

And again $e^{s\ln(x)}$ is an oddly written $x^s$, unless you have a specific reason. Then you should tell it.

geocalc33

1:26 PM

so it is closed :P

Rudi_Birnbaum

@geocalc33 No it isnt. When you claim adding your equations is a group operation you have to show that adding $x^{s_1} + (1-y)^{s_1} = 1 $ to $x^{s_2} + (1-y)^{s_2} = 1 $ gives you $x^{s_3} + (1-y)^{s_3} = 1 $ for some specific $s_3$. So you really should be able to rewrite $x^{s_1} + (1-y)^{s_1} + x^{s_2} + (1-y)^{s_2} = 1+1$ into $x^{s_3} + (1-y)^{s_3} = 1 $ and nothing else.

So you essentially have to give $s_3$ ans a function of $s_1$ and $s_2$ such that the equivalence of the two equations hold.

rschwieb

@MatheinBoulomenos Morning. I have another commutative algebra question for you, when you have time.

Rudi_Birnbaum

@geocalc33 And anyway how do you know that any of these equations make sense? E.g. if $s=0$ then you have $1+1=1$ that does not hold (no matter for which $x,y$)... Yeah maybe in that sense its makes sense adding nonsensical equations will give you a nonsensical equation. Yeah that's closed ...

Alessandro Codenotti

@rschwieb It's very unlikely I can answer, but I'd be interested in hearing the question too!

Sanity check: The proof that of the hairy ball theorem can actually be modified to show that whenever $n$ is even there is no continuous map $f\colon S^n\to\Bbb R^{n+1}$ such that $x$ and $f(x)$ are linearly independent for all $x\in S^n$, right?

Rudi_Birnbaum

1:47 PM

@geocalc33 and even if you manage to fix that somehow (using modified equations) you will never get anything interesting out of it. If you go to linear equations you will end with linear algebra or well known groups like $(\Bbb R, +)$ and the like ...

rschwieb

@AlessandroCodenotti Sure. I can't ask very sophisticated questions anyhow. The question is about the definition of almost-Dedekind domains (and Dedekind domains indirectly.) I know that some commutative algebra texts have the habit of excluding fields from their definitions.

@AlessandroCodenotti For example, the definition of almost-Dedekind is "a domain where the localizations are discrete valuation rings" But at that rate, I guess a field can't be almost-dedekind

So i'm wondering how problematic it is to relax such definitions to "a discrete valuation ring or a field."

Alessandro Codenotti

I don't know about almost-Dedekind domains, but I encountered this in Milne's ANT notes just the other day "According to the above definition, a field is a Dedekind domain. In future, we shall exclude fields from being Dedekind domains (conventions vary)."

rschwieb

@AlessandroCodenotti Yeah sigh. I understand that it eliminates the need for some edge cases, but I really want to know if it's still consistent to be inclusive. In other situations, it usually pays off (in terms of hierarchical classification) to be inclusive.

You can say that a Dedekind domain is just a commutative hereditary domain, but it'd be stupid to say "except if it's a field."

Alessandro Codenotti

Hmmm, i don't immediately see anything that goes wrong without excluding fields

Of course all the stuff about unique factorization of ideals, integral domain with finitely many prime ideals is Dedekind iff pid and the usual results are trivial for fields

rschwieb

@AlessandroCodenotti I guess the question is what happens when you have a domain which has all prime localiztaions that turn out to be fields or discrete valuation rings. Is that even possible?

I mean, to have both

Alessandro Codenotti

2:00 PM

I think the only localization of a domain that is a field is the one at $\{0\}$

rschwieb

OK, well that woudl be good

well

Semiclassical

@Rudi_Birnbaum a non-solution in search of a problem

rschwieb

Sorry, taking some time to process. I'm not used to thinking about domains really. I know that von Neumann regular rings have localizations that are all fields

but VNR's are very far from domains :)

Alessandro Codenotti

If you localize an integral domain $A$ at $\mathfrak p$ you get a local ring with $\mathfrak pA_{\mathfrak p}$ as unique maximal ideal, so it is not a field, or am I missing something?

rschwieb

If I localize at a (nonzero) maximal ideal and the result is a field, that means that the maximal ideal is a minimal prime, but that won't fly if there's the zero prime sitting underneath

so I think I agree :)

Yeah I think we're saying the same thing

Alessandro Codenotti

2:05 PM

Hmm I'm not familiar with your terminology so I'm not sure about your argument, but I'm glad we agree :)

Rudi_Birnbaum

@Semiclassical :-)

Alessandro Codenotti

@AlessandroCodenotti Actually here is enough to have a commutative ring $A$, no need for a domain

user131753

@TobiasKildetoft Wait. I am not sure I understand that comment. Can you clarify a bit?

rschwieb

@AlessandroCodenotti Well, you can definitely have a ring whose localizations are all fields and yet the result isn't a field.

Alessandro Codenotti

Really? I must be missing something, how do you do that?

Leaky Nun

2:08 PM

@AlessandroCodenotti $\Bbb Z/6\Bbb Z$

note that the map to the localization doesn't need to be injective

loch

or easier try product of fields :)

rschwieb

@AlessandroCodenotti Any product of fields at all, even infinite!

@LeakyNun Heya! I put up the ring maps this morning.

They're kind of hidden though

And I accidentally included a suboptimial version of the full map

ringtheory.herokuapp.com/maps

Alessandro Codenotti

Hmmm, What's the localization of $A=\Bbb Z/6\Bbb Z$ at $\mathfrak p=(2)$?

rschwieb

$\{1, 5\}$ were originally units, and then $\{3\}$ gets thrown in, so you have the field of three elements.

Might take a bit to write down the actual equivalence classes :)

Alessandro Codenotti

Hmmm, I'm missing something, let me think about this for a while

loch

2:22 PM

@AlessandroCodenotti i think this follows from the corollary mentioned in en.wikipedia.org/wiki/Hairy_ball_theorem#Corollary

Abdelrhman Fawzy

From Spivak's book Question:

Find a $\delta$ such that $|f(x)-L|<\epsilon$ for all x satisfying $0<|x-a|<\delta$

$$f(x)=\frac{1}{x};a=1 ,L=1$$
we need to find$\delta$ such that $|\frac{1}{x}-1|<\epsilon$

Alessandro Codenotti

@loch looks like it, thanks

loch

actually I think you expect to get the field of two elements (unless im being dumb)

you can use the fact localization commutes with quotients, so $A_{\mathfrak{p}} = \mathbb{Z}_{(2)} / (6)\mathbb{Z}_{(2)} = \mathbb{Z}_{(2)} / 2 \mathbb{Z}_{(2)} = \mathbb{F}_2$.

Alessandro Codenotti

I'd also expect $\Bbb F_2$ because then $(2)$ is the $0$ ideal in the localization, so the unique maximal ideal as expected

rschwieb

hmmm

Alessandro Codenotti

2:32 PM

My argument above should still work for integral domains though because there $A\hookrightarrow A_{\mathfrak{p}}$ is injective and $\mathfrak pA_{\mathfrak p}$ is a nonzero proper maximal ideal, hence $A_{\mathfrak p}$ can't be a field

rschwieb

Yes, it does work for integral domains.

Yeah, it does look like the field of two elements, judging from the equivalence classes... why does inverting 3 (formerly an idempotent) not add a new unit?!

(1,1), (3,3), (5,5) are all obviously equivalent.
Then (1,3)(3,3)=(3,3)
(1,5)(3,3)=(3,3)
(3,5)(3,3)=(3,3)
(5,1)(3,3)=(3,3) etc

thinking of them as fractions

counterintuitive, especially for me

ahh, pff

better not to think of the things in the complement as inverted, at least not when this isn't a domain

rschwieb

2:55 PM

With a quotient ring, you just "coalesce". With a local ring, you somehow have to expand (by looking at R\times S) and then coalesce with the equivalence relation

That makes it a lot harder for me to think of :/

Alessandro Codenotti

It's definitely not intuitive at all for generic rings

With integrals domain you're just working inside $\operatorname{Frac}(A)$, which is super nice

rschwieb

Right

loch

i just realised i completely forgot about chinese remainder theorem
oops

geocalc33

@Semiclassical I don't appreciate your comment. The idea I'm working on is in its infancy and it's not going to be perfect on the first try. It is very important to who I am, and when you imply that I'm fumbling in the dark looking for solutions when there are none to find it's insulting and makes me feel of lesser worth

Perturbative

Hey everyone

So a local ring is defined as a ring $A$ with exactly one maximal ideal

But can't there only ever be one maximal ideal for any ring $R$?

Like if you have two maximal ideals, shouldn't they be equal set theoretically?

loch

3:10 PM

@Perturbative have you thought about $\mathbb{Z}$?

Mike Miller

@geocalc33 It's important to separate your feelings of work from whatever work you do - partly so that there are no hard feelings from things like this, partly because research is hard and miserable.

I have spent a lot of time feeling worthless because I was stuck or couldn't solve something I expected to be able to.

Feelings of worth*

Perturbative

@loch Ahh yes of course, the ideals $(p)$ where $p$ are prime are maximal in $\mathbb{Z}$, but $(p_1) \not\subseteq (p_2)$ and $(p_2) \not\subseteq (p_1)$ for any primes $p_1, p_2$ since $p_1 \not\in (p_2)$ and vice versa

So there's no one real maximal ideal in $\mathbb{Z}$

@MikeMiller On the upside some of the best feelings are to be had when you solve something you didn't expect to be able to

Vrouvrou

3:29 PM

hello, can someone help me on union of connected sets?

Alessandro Codenotti

It's connected as well, if they intersect somewhere

Perturbative

@Vrouvrou So in a topological space $X$, if you have two connected sets $C_1$ and $C_2$, if $C_1 \cap C_2 \neq \emptyset$ then their union is connected

Vrouvrou

3:57 PM

@Perturbative i do a prove and I want to see if it. is correct

Rudi_Birnbaum

Again hanging on some triviality (sry): I have an equivalence relation on a monoid. How to see that $a_1 \sim a_2$ and $b_1\sim b_2$ implies $ a_1 b_1 \sim a_2 b_2$. Intuitively its somehow clear...

Leaky Nun

@Rudi_Birnbaum it depends on your equivalence relation

Rudi_Birnbaum

@LeakyNun No restrictions as far as I can see

Perturbative

@Vrouvrou You can post it here and one of us can check and verify it

Leaky Nun

@Rudi_Birnbaum it's not true in general

Rudi_Birnbaum

3:59 PM

@LeakyNun That relieves me a bit, I still show you the original text 1 sec.

Sry