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5:00 PM
This is soooo non obvious. It appears the mere fact that having just one digit to be switched into 0 partition the sequence into one that is strictly larger and a sequence that is strictly smaller, and this process can repeat indefinitely, thus generating a dense ordering, thus thwart any attempt to well order countable binary strings
 
I think closed under an operation outside the set makes sense
 
okay :D
when we do ring homomorphism btw
 
$N$ is normal in $G$, then $gN=Ng$ for each $g\in G$.
 
the kernel is viewed on (R,+) ?
 
So the fact that the naturals can be well ordered may be more of a fortunate fact that this "bubbling of zero" procedure eventually terminate, thus the partition can only go so far
 
5:01 PM
$I$ is an ideal in $R$, then $rI=Ir$ (commutative ring).
 
because most of the times (R,.) wont be a group
 
@user104729 that's true for every subgroup (if the ring is commutative)
It's just the set of elements that map to zero, so yes, you could say it's the kernel of the additive groups
 
@MatheinBoulomenos Huh?
 
@MatheinBoulomenos okay thanks mathein :)
 
For every subgroup $H$ of $(R,+)$ $rH=Hr$
if $R$ is commutative
so that's not saying much
 
5:04 PM
@MatheinBoulomenos I was talking about $G$ a group, $N$ a normal subgroup of $G$. $G$ is not a ring.
 
@user104729 I responded to this
 
Oh, the group comment confused me.
Really I meant $I$ a two-sided ideal.
 
I dont know why we doing this ><
I mean the ideas are old
 
(For $R$ not commutative necessarily)
 
hello. I wanted to study about nodes in graphs, as in graphs of functions, not in graph theory. But whenever i search for nodes, graph theory pops up. Is there any other name for it?
 
5:05 PM
only fancy names for them
 
@KasmirKhaan The idea of a scheme is old, and to learn it, you'll need the ideas of commutative algebra etc. I.e. you need to learn old things to learn the new things.
 
Hmm I assume it is the proper way for things to come
 
It's just a basic object that comes up all the time.
 
Then kasmir shall learn it :D
Okay thanks yall for help
Ill keep working =p
 
Why are we doing this? An important question to ask. The reason to care about rings and ideals is that a lot of objects we care about have the structure of rings. Integers (and in algebraic number theory we see, that we can learn a lot about integers by extending them to larger rings), polynomials and several kinds of functions on a "space", whatever space means
The ideals of a ring tells us something about the structure of the ring, so we can understand the ring better if we look at the ideals
 
5:08 PM
@MatheinBoulomenos I see :) this is why I like to ask you :D you give me good understanding of why :D
I have to understand the isomorphism theoms and how to use them
 
@LeakyNun It took me a while, but I found the infinitely decreasing chain
X denote entries that are made zero
{1,3,5,7,9,11,13,15,17,19,21,23,25,...}
{X,3,5,7,9,11,13,15,17,19,21,23,25,...}
{1,X,5,7,9,11,13,15,17,19,21,23,25,...}
{1,3,X,7,9,11,13,15,17,19,21,23,25,...}
{1,3,5,X,9,11,13,15,17,19,21,23,25,...}
...

There are uncountably many of these. Basically, take any countable sequence and then take turn omit one entry in order. Then to the left of the omitted entry is a subsequence and to the right is another subsequence strictly larger than that on the left. i.e.
 
we did not spend much time on them on Groups and then we used them on rings like we knew them
><
 
nice @Secret
 
In particular you'll consider something more general than vector spaces, called modules. You can treat a ring as a module over itself, and the ideals become submodules. If the ring isn't commutative, the left and right ideals will become left and right submodules respectively.
 
It extremely nontrivial, and I am still trying to understand how it happened
 
5:10 PM
(Where the vector space is a module over a field, and a module is simply over a ring)
 
you're essentially giving the ordering of the p-adic integers @Secret
 
@user104729 oh ill keep that in mind :D i Heard the Word module but dont know what it is yet =p
 
consider ....1110, ...1101, ...1011, ...
that's basically what you are doing
 
@KasmirKhaan Just an ideal in a sense :P.
 
it seems like it has rich structure
 
5:11 PM
Abelian group under addition, closed under 'scalar multiplication' from a ring (if the ring is a field it is a vector space).
 
but for me atm, that is just empty Words << so i need to do the basics properly
 
Sure.
 
Well I'll be afk again for some time. Good luck for now!
 
thanks ! see you next time :)
 
5:13 PM
As an exercise, try to show that a (edit: commutative) ring R is field if and only if there are exactly the two ideals {0} and R
this is an example of how the ideal structure tells you something about the ring
 
hmm let me think
a field is a commutative division ring right?
all elemnts are units exept for 0
 
Yes
I assumed that R is commutative
 
oh yes yes
hmm, if we assume R is a field
and R is not {0}
 
ugh, I really need to figure out how to formally prove that any attempt to well order an uncountable set explicitly will result in a contradiction by producing infinitely decreasing chains
 
since 1 is in I,
then a in R ==> a in I
so by multiplication with 1 we get all the ring R
 
5:18 PM
why is 1 in I?
 
because we have a field no?
how can elements be invertible if we dont have 1
 
Yes, you have to use that
 
hmm let me think
 
I was assuming that I is an arbitrary nonzero ideal
 
well if a in I
 
5:19 PM
@KasmirKhaan that's what you get for not letting your variables
"since 1 is in I"
 
I think any well ordering of an uncountable set cannot be expressible in the algebra of strings, for the information on the ordering is prescribed by the content of the string (which can always be produced by an infinite time turing machine).
Therefore, for any countable string, one can repeat the "zero bubbling" game by moving zero to the next largest position in the string and then this will guarentee partition of the string into two substrings where one is larger than the other in the ordering
 
should be preceded by "Let I be an ideal."
then you will know why you can't assume 1 in I a priori
 
I dont get it leaky
but if we start this way
let a in I,
 
@KasmirKhaan ...
 
You have to define I first
 
5:20 PM
what is I?
 
since we have a field, there exist a'
let I be nontrivial ideal
a in I
 
why do you know a in I?
nontrivial ideal?
 
leaky help me get foward not get back
let me finish my statement :D
 
it's more common to say nonzero, although nontrivial will probably be understood
 
okay
I is a nonzero ideal
now it make sense to have an element a in I
since we have a field, there exist a'
 
5:22 PM
@KasmirKhaan no, you need to say that a is non-zero
 
ok non zero element a
 
 
now kasmir lost where he was going to with this
thanks leaky
 
someone can explain to me why "it follows" ?
 
-.-
all right i Think i got it
 
5:24 PM
it is in measure theory, proving that the lebesgue measure is invariant
 
aa'=1 so 1 in I
since we got 1 in I, we can have all elements of R in I
this I =R
 
@Liad well "sets of measure zero are preserved by translation and dilation" is just a restatement of "if m(E)=0 then m(E+s) = m(rE) = 0"
 
@MatheinBoulomenos well? :)
 
3 mins ago, by Kasmir Khaan
leaky help me get foward not get back
 
but we know it only for borel sets@LeakyNun
 
5:25 PM
you pull yourself back with your lack of rigor
 
what happens if $a=0$?
 
he wants to prove it for every lebesgue-null set @LeakyNun
 
leaky ! am not on exam now,i just want to see if i got the important idea right
 
@Liad well I presume author is talking about the real Lebesgue measure?
 
the idea is right
 
5:26 PM
well if a=0 , then I={0}
 
which is only defined for Borel sets?
 
thanks :D
 
it is not defied only for borel sets
 
@KasmirKhaan the idea is right but you need more rigor
 
@Liad What is your measure space?
 
5:26 PM
that comes with experience leaky
 
@user104729 $\Bbb R$
 
Iam alot better now compared to 2 months ago
 
@Liad how do you define a Lebesgue-null set?
 
so things take time
ill make sure to define Everything properly on exam
 
that's the question
i know that every lebesgue set is the union of borel set and a lebesque-null set
 
5:28 PM
@KasmirKhaan I hope so
@Liad but what is a lebesgue-null set?
 
how is it defined in your book?
 
@Liad I think it's an error.
 
@user104729 what is?
 
They are establishing that this holds for $\mathcal{B}_{\Bbb R}$ at this point, and then using this for the $\mathcal{L}$ case.
 
5:30 PM
alright, what's wrong with that?
 
I just mean that the sentence "from which it follows..."
It should say that these properties hold for $\mathcal{B}_{\Bbb R}$, rather than Lebesgue measure zero sets.
Then it should proceed as it does. Do you agree with that?
 
so what's lebesque null set ?
 
how does your book define it? there are multiple equivalent definitions
 
im a bit confused to be honest :P
 
if you have established that $\mu (A) = \sup \{ \mu (F) \mid F \subseteq A, F \text{ compact and measurable} \}$, then I see how it follows
 
5:36 PM
we didn't
we saw $\mu(A) = inf \{ \mu(U) : A\subset U\subset \Bbb R $ is open$\}$
 
from this you can conclude that Lebesgue null sets are preserved under translation and scaling if you know how the measure behaves under these operations for Borel sets
But you don't need the intermediate step that $A$ Borel and $\mu(A) = 0$ implies $\mu(A+x)=0$ and $\mu(rA) = 0$
 
@AlessandroCodenotti i am now but i have been studying physics
if you have a question, shoot
 
Hello. i have to show that all real valued solutions of $y'' + siny = cosx$ exist for all $x$. Please give me a hint, how to procede
 
@MatheinBoulomenos alright i think i got it. thanks.
 
anyone?
 
5:59 PM
$\infty - \infty = \pi$
 
@NV-US what does that statement even mean
 
@LeakyNun what's your view?
 
@Abcd on what?
 
@LeakyNun on the paradox obviously..
 
I don't see any paradox
as long as you define everything clearly
 
6:02 PM
@LeakyNun do you agree with $\infty- \infty = \pi$?
 
Is it not intuitively obvious that $\infty-\infty =\pi$?
 
@Abcd not really
 
@user104729 No.
 
define infinity
 
@Abcd is the shouting necessary?
 
6:03 PM
@LeakyNun Do you agree with things like $-1+1-1+1...= 0.5$?
 
Wait, I thought $\infty = -\frac{1}{12}$
 
@MatheinBoulomenos That's why I thought it held too.
 
@LeakyNun watching the video, will let u know
 
@LeakyNun Not really "shouting" but capitalised for emphasis.
 
@Abcd i.e. shouting
@Abcd I do not
 
6:05 PM
@LeakyNun what did u not understand in my statement?
 
"paradox" does not have to mean "self-contradictory", it can also mean "seemingly absurd"
 
@NV-US the whole statement
 
contradictions are absurd
 
all real valued solutions of $y'' + siny = cosx$ are continous on $R$ @LeakyNun
 
in Mathworks (Not the main chat!), Oct 3 at 15:25, by user21820
Let me explain what is the real issue here. You may be surprised when I say it has nothing to do with convergence or zeta functions.
@Abcd my stance is mostly the same as user21820's stance, so please read the conversation there
 
6:07 PM
I was wrapping my head around the Zariski topology on $\text{Spec} R$ @Balarka
 
highlights:
in Mathworks (Not the main chat!), Oct 3 at 15:29, by user21820
If that person cannot define what on earth he means by 1+2+3+..., then he/she has no right to claim it even exists..
 
@AlessandroCodenotti Is there something that troubles you in particular?
 
in Mathworks (Not the main chat!), Oct 3 at 15:29, by user21820
If that person is interested to know what mathematicians conventionally define it to mean, then it's time to explain real analysis.
 
Not in particular, I'm trying to build some intuition
 
6 mins ago, by Abcd
@LeakyNun Do you agree with things like $-1+1-1+1...= 0.5$?
In particular, I do not think that you know how your statement is even defined @Abcd
 
6:09 PM
Why the ad hominem?
 
@MatheinBoulomenos how is it an ad hominem?
 
How is "I do not hink that you know how your statement is even defined" not an ad hominem?
 
@LeakyNun What do you mean?
 
@AlessandroCodenotti can u help me with my question?
 
@MatheinBoulomenos because it isn't an attack?
 
6:11 PM
So, let's do an easy example, $\Bbb R$. Maximal ideals in $\Bbb R[X]$ are those of the form $(X-r)$ for some $r\in\Bbb R$, so the closed sets in the Zariski topology on $\Bbb R$ are finite sets of points and the Zariski topology on $\Bbb R$ should agree with the cofinite one, is that right?
 
@MatheinBoulomenos I'm merely stating that I do not think he knows how the expression is defined
therefore it is meaningless to ask me whether I agree with the statement
because there is no statement to begin with
 
@AlessandroCodenotti you forgot about $(0)$
 
how is this an attack?
@Abcd I mean it is meaningless to ask whether I agree with that statement before you have even defined that expression precisely
 
@MatheinBoulomenos Ah, of course, $(0)$ is prime
 
@LeakyNun people say that without stating any definitions..
 
6:13 PM
(0) is prime?
 
in every integral domain
 
@Abcd ok?
 
actually a commutative ring $R$ is an integral domain iff $(0)$ is prime
 
then be the first one to make a change lol @Abcd
@AlessandroCodenotti surely, because R/M is integral iff M is prime
and R/(0) = R
 
@LeakyNun Where is the directrix of the circle?
 
So $\operatorname{Spec}(\Bbb R[x])$ is $\Bbb R$ with the cofinite topology and with additional point which happens to be dense
 
> So, to answer your question: As its eccentricity is zero, a circle doesn't have a defined directrix in a two dimensional plane.
 
@MatheinBoulomenos Wait why is it dense
 
that's a good exercise
Basically, it's because $(0)$ is contained in every prime ideal
 
@LeakyNun Someone told me it's at infinity and it has been proved mathematically.
 
6:19 PM
@Abcd that may be a useful intuition
but it holds no mathematical truth
(inb4 RP2 bs) (amirite @MatheinBoulomenos)
oh, show me the proof then
 
@LeakyNun That person said there's a proof, I don't know.
 
So projective geometry is bs? that's abold statement to make
2
This is only a matter of definition if you prefer "undefined" or "infinity"
 
@Abcd it may be true in some definition of the real plane, but in the usual real plane which does not include infinity, that statement holds no truth
other definitions have points and even lines at infinity
but we call them "projective plane"
 
If $p$ is a prime in some integral domain, $n \in \Bbb{N}$, and $a$ some element such that $a \mid p^n$. Is it true that $a = p^k$ for some natural number $k \le n$?
 
@AlessandroCodenotti I guess here $\Bbb R =\Bbb C$? Else you are missing many things.
 
6:22 PM
Nah, we don't necessarily need an algebraically closed field
or even a field
 
@user193319 I think so, because prime is irreducible
 
I mean that you missed many prime ideals :P.
 
Oh yeah, how did I not see this
 
If you meant $\Bbb{C}[X]$, then you were right.
 
Ah, right, there's more irreducible polynomials
 
6:23 PM
You said $\Bbb{R}[X]$.
Yep.
 
@LeakyNun In $\Bbb R^2$, there is nothing at infinity. In $\Bbb R\rm P^2$, there is a line at infinity. (To use the proper notation.)
 
@LeakyNun Sure. That's right: primes are irreducible. But why does this imply that $a \neq p^k$? We know the theorem is true for integers.
 
@user193319 did I say $a \ne p^k$?
 
And these are both different from $\widehat{\Bbb R^2}=\Bbb R^2\cup\{\infty\}$.
 
sure
but in the setting of @Abcd's question, I'll go with that he is actually talking about $\Bbb R^2$, in which case the statement is clearly false
 
6:25 PM
@LeakyNun Oh. I misread your first reply. Sorry about that. I'll try to think about the proof some more.
 
hmm, actually I'm confused, if I have a generic ring $R$ the Zariski topology on $\text{Spec} R$ has as closed sets the sets $V(I)=\{P\in\text{Spec}|P\supseteq I\}$, for all ideals $I$ of $R$
 
@user193319 $pp^{n-1}=ab$ for some $b$, so $p|a$ or $p|b$. Proceed by induction.
 
@AlessandroCodenotti What confuses you with this?
 
Ah, ok, so in the example with $\Bbb R[X]$ the smallest closed set containing $(0)$ is the whole space, which explains why its dense
 
Yes, it's a generic point.
 
6:28 PM
generic?
 
Such points are called generic points.
 
The name suggests that they are pretty common, is that the case?
 
The reason for the name is here fortunately: en.wikipedia.org/wiki/Generic_point
 
Ah, I see
So in an integral domain $(0)$ is generic
 
Yep.
The unique minimal prime ideal.
 
6:32 PM
And it's also the only generic point since $\sqrt{(0)}=(0)$ and that's the intersection of the prime ideals
 
donuts are good for you
yum
 
Dense points are bad
@Alessandro It's generic in the sense that, if $X$ is a subset of $Y$, points from $X$ are generic in $Y$ if $Y \setminus X$ is in some sense "thin" or "sparse"
Eg set of regular values of a smooth map
In this case, the point is dense is the space, so the complement is kinda sparse :P
It's pretty dumb
 
I shouldn't have said anything :P.
I induced a random tangent :P.
 
I see, thanks!
 
I mean I just pulled that explanation out of my [redacted for using non-christian word] anyway
 
6:42 PM
It's dinner time now, I might have more questions later
 
While you eat, just think about the fact that some variety that isn't one dimensional corresponding to the ideal $I$, has a $V(I)$ that contains the maximal ideals, corresponding to all of the points on this variety.
(For intuition)
 
@LeakyNun I'm having a little trouble proving the claim. If $a \mid p$, then $p = ab$ for some $b$, and so $p \mid ab$ which implies $p \mid a$ or $p \mid b$. Here's where I get stuck. Even if I could show that $p \mid a$, this would only show that $a$ and $p$ are associates and therefore $a = up$ for some $u$. But I don't know how to argue $u=1$. Even if I could do that, I'd still have to deal with the $p \mid b$ case.
 
@user193319 let $a=up$, then $p^{n-1} = ub$, induction.
oh right, $a$ may not be $p^k$ itself, but an associate of $p^k$
 
Spec R is slightly "finer" than maxSpec R, being the upshot, I suppose
's all a bullshit topology anyway
 
why would $p|a$ mean $p$ and $a$ are associates? @user193319
 
6:46 PM
Never thought I'd hear Balarka say the words "bullshit topology" in my life
 
HAHAH
r/nocontext spotted
 
@LeakyNun Because $a \mid p$ and $p \mid a$ implies they are. You told me to do induction, so I took $n=1$ as my base case.
 
alright
 
@BalarkaSen >:(
 
To be fair, I didn't actually hear him say this since he only has fingers, not a mouth, but still
 
6:47 PM
@LeakyNun So might the claim be false?
 
@Daminark I have a finger, I have a keyboard.... UHHH: word salad
 
@user193319 eh, are you sure the claim did not include "associate"
 
Kek
 
No, because I made the claim up myself.
 
alright
then $-2 \mid 4$ but $-2 \ne 2^k$ lol
 
6:49 PM
But some claim involving the word "associate" is probably true. This may be all I need to prove another problem I am working on.
 
@user193319 yes
 
Neat. Thanks!
 
Finally got the last algebra pset
 
how many questions?
 
10 questions, which is pretty standard fare
 
7:01 PM
any proofs?
 
Mostly, and even the examples require proof
 
cool, thanks
 
I am interested to know how fast the "Dirichelet Eigenvalues", explained on the wikipedia page (en.wikipedia.org/wiki/Dirichlet_eigenvalue) go to infinity. More specificially, I am wonder if the series consisting of the reciprocals of the eigenvalues is in \ell_1. What words do I look for online?
 
hello @Daminark
 
How's it going?
 
7:06 PM
good
how are you?
 
Quite busy, but aight
 
oh
continue
 
7:29 PM
How to show that if $B$ invertible then trace $A$=trace $BAB^{-1}$
 
@Silent do we know trace AB = trace BA?
 
@MikeMiller hi
 
Yes@LeakyNun, but I can't see how this follows from that, because, e g, trace $AB$= trace $A$$\cdot$trace $B$ does not hold.
 
@Silent tr (B)(AB’) = tr (AB’)(B)
 
To answer my own question, I needed to look up "Weyl Assomptatics".
asymptotics.
 
7:36 PM
@LeakyNun Wow! I should have thought a little more. Thank you so much.
 
Morning
 
@LeakyNun i need some help with a logic question, i have $A=\{ 1 + 1/n\} , B = \{1 + (-1) \ ^ i / n :i \in \{1,2 \} \} $. i only have '<' and need to separate $A,B$ with a formula $\phi$
(for example we can separate $\Bbb Q$ and $\Bbb Z$ with $\phi = x<y$ exists $r (x<r)$ and $r < y$ )
 
7:52 PM
How do you deal with someone who won't accept that their proof is a crank proof? lol
 
@ÍgjøgnumMeg You don't. You move on :P.
 
He keeps trying to make me review his "proof" of the Collatz conjecture lol
 
wth is a crank proof
 
@Faust Using delta brainwave chains and the axiom of 'if there is a beginning, there is an end', to prove the Riemann Hypothesis.
 
8:09 PM
O.o
 
8:20 PM
The Riemann Hypothesis has been solved about 15 times on ArXiv
 
@Liad $\exists x \forall y [x < y]$
 
@LeakyNun huh ? take $2$
 
what about it?
 
wait i read it in the opposite direction :P
 
I have to go and sleep now. But if someone else can find it, I was looking for a paper that proves a big conjecture in a very cranky way. They have brain waves and delta-chains, and they have a few axioms that are stuff like "Treat thy neighbor right" and "Everything that has a beginning has an end".
 
8:25 PM
they are both bounded from below so i dont see how this works @LeakyNun
 
@Liad doesn't mean they both have minimum
 
@LeakyNun do you see why my formula does not work?
 
does your verifier check for names?
 
what do you mean?
 
8:30 PM
the "and" should be "implies" anyway
 
i did it wih implies too, did not work
 
I mean how does your verifier know if your answer is correct
does it count Square and SQUARE different? can I just type sq?
 
you just click on "Square" and fill in $x$
 
can you show me?
 
 
8:33 PM
then you left out triangle
 
Huh?
small triangles
 
it says all the small triangles
I don't see triangle anywhere in your answer
 
Ops
 
imo all translations questions in logic should be done in lojban xd
 
8:47 PM
@LeakyNun 0 is minimum of both A and B
 
@Liad $0$ isn't even there
 
$\{1-1/n \}$ . take $n=1$
 
that isn't what you wrote to me about A
 
My bad. sorry, tired :/
 
...
 
8:49 PM
so $A =\{1-1/n \} B = \{1 + (-1 ) \ ^ i /n \}$
 
$\exists x \forall y [y<x \lor y = x]$
 
hm. 2 would work for both , doesnt it?
 
what is 2?
 
take $x=2$
 
it isn't in A
 
8:52 PM
to prove the statement $dim(image) = dim(preimage) - dim(kernel)$, is it sufficient to show that if some nonzero element $e \in V$ where V is the preimage vector space, is mapped to zero, then by linearity all elements colinear with $e$ are mapped to zero, and then any other element can be broken into $A(p) = A(q + ce) = A(q)$, which is a projection, and that subsequently the number of lines that lie in the kernel that aren't expressible in terms of the others correspond to...
The dimension of the Kernel, and the dimensions lost from the Preimage
 
you right, i thought we look at them as a subset of $\Bbb R$
 
@Phase what is "preimage"?
do you mean domain?
 
why 1.5 wont work for $B$ ?
 
yeah I guess
 
@Liad because there's 2
 
8:56 PM
how so?
 
$1+(-1)^2/1 = 2$
 
which $n$ will give you $2$
Ok. so why 2 wont work?
 
it does
 
and 1 works for A
 
it isn't there
 
8:58 PM
right!! same mistake :/
 
it helps to draw your set out
 
i will, thanks.
 

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