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12:04 AM
Hello. I have a probably trivial commutative algebra question. If $R$ is any commutative ring, why do we have $\dim R[x]\geq \dim R+1$? When $R$ is not a domain the ideal $(x)\vartriangleleft R[x]$ may not be prime (take e.g $R=\mathbb Z/4$ and $x+2\in \mathbb Z/4[x]$).
 
Hi guys
may someone tell me if my proof on that every subset of a discrete metric space is open, is correct, please?
Case 1) If A is the empty set, then A is open.

Case 2) Let A be a non empty subset of the $X_{discrete}$ . Let a$\in$A and set r$=$ $\frac{1}{2}$ then if x$\in$ $N_r(a)$ then $d(x,a)=0$ and so $x=a$, hence x$\in$A.
 
12:20 AM
Mathematics is the art of generalizing phenomenons and concepts. One of the most representative symbols in mathematics is the letter x, and it perfectly encapsulates the idea of generalization: we might observe the area of a square with side length 3 to be 3 times 3, and the area of a square with side length 5 to be 5 times 5, so we might extrapolate and claim that the area of any square is its side length times itself, or in short: the area of a square with side length x is x times x, demonstrating how the symbol x is related to generalization.
@mathsssislife why divide into two cases when your proof of case 2 did not rely on the non-emptiness?
the art of generalizing phenomenons and concepts -> the art of generalizing patterns
 
I know phenomenons is an accepted plural but
 
oh no how did I miss that lol
no I just forgot
 
lol just making a joke anyway
 
anyway @ÍgjøgnumMeg I'm in love with modular representation theory
 
What is it?
if you wanna go ahead and give me a summary
lol
 
12:26 AM
it's representation theory when the characteristic of the field divides the order of the group
 
I see
 
but I guess I've only done one problem sheet so maybe I shouldn't say it so early lol
 
@LeakyNun what do you mean by that my proof of case 2 did not rely on non-emptiness? I assumed it is non empty,and since it is non empty it has an element.
 
then perhaps your proof is wrong to start with
 
Why? can't a subset of a metric space be empty?
 
12:29 AM
a set A is open if for every a in A there is r such that B(a,r) is contained in A
you just proved it for a specific a in A
 
ohhhh, I see, the empty set is an element of every set
 
no it isn't
 
Well, it's an element of every power set, but a set of non-set objects shouldn't contain a set.
 
Couple of things I wanna focus on in my free time: more topology, representation theory, continuing with ANT (local fields)
 
non-set objects, would I be incorrect to say that the empty set is an element of every metric space?
 
12:33 AM
@ÍgjøgnumMeg and what do you do in your non-free time?
@mathsssislife yes
 
@Leaky I work in the IT department of my university, have an interview on tuesday to work in a company developing mathematics education software
 
wait what?
that's a huge plot twist
 
@LeakyNun what would be an example of a metric space that doesn't have the empty set?
 
@mathsssislife well take the real numbers for example
 
12:34 AM
@mathsssislife the empty set is a set, it's a subset of every set, if your set is not a set of sets then it can't have elements that are sets
 
@ÍgjøgnumMeg well I thought you're math undergraduate, but maybe I just assume everyone is a copy of myself by default
 
@Leaky I was a math undergrad, I graduated in September 2018, was supposed to go to Heidelberg this year for my masters but deterioration in my mental health meant that I failed to get my application in on time l o l
 
@ÍgjøgnumMeg but every set is a set of sets
 
oh...
 
@Alessandro pls don't
 
12:36 AM
@AlessandroCodenotti spotted the set theorist :P
 
the worst part is that the admissions people said my application was essentially perfect, just that it was late
 
maybe try next year?
 
My plan is thus; if my interview on tuesday goes well and I get the job then I'll apply in March 2020 (if I enjoy the job) so that I can start in the summer semester 2020
this means I can save money for the time, and do some more self-studying (I have no problem with this anyway, all of my pure maths knowledge is self-studied)
 
@LeakyNun I see what you mean, since by definition a set is open if "for every" then I shouldn't divide by cases, for emptiness and non-emptiness.
 
if I DON'T get the job, I'll make a quick application to Frankfurt am Main and hope I get that and then just rush over to start hahaha
 
12:39 AM
I think your last remark applies to me also :P (the self-studied part)
 
Also, at some point I might stop moaning about it on this chat
 
I have a very difficult time going to lectures consistently
 
maybe
 
@ÍgjøgnumMeg Hope it goes well brother, you deserve the best
 
@ÍgjøgnumMeg I'm sorry to hear about your application :(
Hope all goes well for you!
 
12:40 AM
@Leaky at least you have pure stuff available to you, almost my entire curriculum was statistics and applied maths, both of which are entirely mystical to me (rendering the majority of my degree useless in my eyes, though not in the eyes of employers, but who cares about them?)
Thanks y'aaaaaall
 
@ÍgjøgnumMeg how old are you if I may ask ? and where are you from ?
 
@LeakyNun so all I should have said was: Let A be a subset of $X_{discrete}$ . Let a$\in$A and set r$=$ $\frac{1}{2}$ then if x$\in$ $N_r(a)$ then $d(x,a)=0$ and so $x=a$, hence x$\in$A. am I right?
 
@jacksoja 24, United Kingdom of ProbablynotintheEUnextyear
 
@ÍgjøgnumMeg you still young and have a lot of opportunities so dont worry about this
I did not follow brexit well, so am not sure if they leave EU is good or bad
 
@Jacksoja cheers :) if I'm honest I don't follow it particularly well, but I identify completely as European rather than British, so it's like being torn from your country because some fat old racist white men decided it's the best idea
Excuse me if I cause offence (but also not if you voted leave)
 
12:44 AM
@ÍgjøgnumMeg Let $G$ be a finite group, $K$ be a field, define $I_G := \{ \sum_{g \in G} \lambda_g g \mid \sum_{g \in G} \lambda_g = 0 \}$, show that $I_G$ is generated as a $K$-vector space by $\{g-1 \mid g \in G\}$
or not
 
@ÍgjøgnumMeg haha I never heard this part, I thought the ppl wanted brexit
well media always give us what they want to give us I suppose
 
Hey Handsome ppl
 
looks away
 
So I got a question that is rather silly but
 
looks directly into mirror
 
12:47 AM
@LeakyNun only Ted can pull that move
@ÍgjøgnumMeg stop acting like Ted -.-
okay so if we have a field K
and a subfield L
 
@Leaky if this requires tools from rep theory then i have no such kit
lol
 
it is qute normal that K is a vs over the smaller field L
 
@ÍgjøgnumMeg it doesn't require any such tool
 
but how to find the dimension of the vs K ?
if we assume this example
K = Q and sqrt 2 and sqrt 3
 
I object to your choice of variables
 
12:49 AM
@Kasmir identify $K$ with $L[X]/(f)$ where $f$ is irreducible over $L$ I guess
 
L extends K
 
err
$L[X]$
 
@KasmirKhaan for your case use the tower law
 
Okay but this is rahter handsome tbh
never thought of numbers in this way
 
12:50 AM
find $[\Bbb Q(\sqrt2,\sqrt3):\Bbb Q(\sqrt2)]$ and $[\Bbb Q(\sqrt2):\Bbb Q]$
 
we can find a basis for K
1, sqrt 2 , etc
but no idea how to show that they are lin indep
I mean I know by trial an error
 
here's an answer I wrote recently
 
okay let me see :D
@ÍgjøgnumMeg thanks !
 
Can you check my question that i have a bounty on ?
 
12:52 AM
@KasmirKhaan for a high tech way... maybe I should stop talking before Ted smacks me
 
if you link it here others might also take a look
 
al righty , then i shall do that:D
5
Q: Finding a homomorphism between groups with a given kernel

Kasmir KhaanWhat is a homomorphism defined on the group of invertible upper-triangular $3\times 3$ matrices whose kernel consists of matrices $\begin{bmatrix} 1 & 0 & a \\ 0 & 1 & 0 \\ 0 &0 & 1\end{bmatrix}$? I want to use this to study the quotient group, also is there always a way to find a group homomorp...

the bounty goes off in one day:D
 
Guys, are these two statements equivalent: A set U is open if for every u$\in$ U $\exists$ $r>0$ such that $N_r(a)$ $\subset$ U , A set U is open if $\forall$ u$\in$U $\exists$ $r>0$ such that if for every x $\in$ $N_r(a)$ then x$\in$ U. Also, If I just the remove the $\textbf{for every}$ x $\in$ $N_r(a)$ does that still mean i'm universally quantifiying over x?
 
@LeakyNun are you one of those guys that do the "close duplicate " things?
 
no
 
12:56 AM
@LeakyNun so you are not on that list?
-.-
 
well I interpreted "do" as the habitual present
 
I knew it :D
 
as in, whether I actively find questions to close as duplicates in my free time
maybe you should have phrased your question better
 
Haha no not like that
It was for fun nothing else
Okay yall thanks for help
Ill come back later with more Q's
 
@Leaky have you ever looked at any Iwasawa theory?
 
1:04 AM
no
 
it's interesting, though I lack the tools to understand anything but statements lol
Kummer's proof of Fermat's last theorem fails for irregular primes because his proof depends on the fact that the class number of the $p$th cyclotomic field (for $p$ regular) is prime to $p$, classical Iwasawa theory looks at the behaviour of irregular $p$ and why Kummer's proof fails at those primes
 
do you agree that definitions are less important than properties and examples?
 
by looking at $\Bbb Z_p$-extensions of $\Bbb Q$ I guess
 
my friend asked me what an algebra is in the context of algebra over a ring
 
I don't think it's fair to say they're less important, you can't have examples without definitions
 
1:08 AM
I hesitated because I was searching for the easiest definition to understand (and he interpreted this as me not knowing the definition)
I guess I'm just holding a grudge :P
 
yeah it's good to gauge the level of a person before answering though so that's the right thing to do
 
you don't think in terms of definitions
if you see a group-theoretic exercise
 
if someone who doesn't know what a group is asks you what a vector space is and you answer by saying it's an abelian group together with an action blah blah then it's not going to help
 
you don't think about the definition of a group
you think in terms of properties and examples
 
I guess so
 
1:13 AM
@ÍgjøgnumMeg I don't even understand the statements
 
What do you have to say about this statement: "Any hypothesis that withstood any test of refutation becomes more likely"?
 
@Leaky lol I understand a few I guess
 
@ÍgjøgnumMeg for a prime $p$, find the JNF of the $p$-by-$p$ matrix $\begin{bmatrix}0&0&0&\cdots&0&1 \\ 1&0&0&\cdots&0&0 \\ 0&1&0&\cdots&0&0 \\ 0&0&1&\cdots&0&0 \\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots \\ 0&0&0&\cdots&1&0 \end{bmatrix}$ over the field $\overline{\Bbb F_p}$
 
1:29 AM
Guys, I know this might be dumb but does every metric space have balls?, because a ball requires a point in the metric space and the metric
 
1:47 AM
You can define balls in any metric space
 
 
3 hours later…
4:29 AM
Guys so I posted this linear algebra question about two days ago; I got an answer but i asked back so i could get some clarification. The person who answered did not respond so I'm asking here. This is the question:
1
A: $A^{C_p} - I_2$ has all entries divisible by $p$, for an infinite number of positive integers $p$

Will JagyI will look for some reference for this: given a nonzero integer $n,$ there are infinitely many primes for which $n$ is a square (a quadratic residue). The two proofs i can think of use quadratic reciprocity, Dirichlet's result on primes in arithmetic progressions, or Chebotarev density. https:...

Does anybody get why the prime can't divide the determinant? Is it because then we're going to have a zero eigenvalue?
 
probably something to do with invertibility, since mod $p$ you'll have zero determinant
 
I thought so but you don't really need the matrix to be invertible
My best guess is that if you have a 0 eigenvalue then you can't have $D^n = I_2$ no matter what n, where D is the diagonal matrix, but I'm not sure that's it
 
You note in the question that $\det A \neq 0$
hang on I'll read the question properly, I'm just taking glimpses
 
 
2 hours later…
6:55 AM
@nbro I have to say that it is either naive or foolish or an attempt to win support for something that doesn't actually have good evidence.
 
7:44 AM
I couldn't understood the proof of bezout's identity (en.wikipedia.org/wiki/B%C3%A9zout%27s_identity#Proof). specifically the statements, "The set S is nonempty since it contains either a or –a". Someone please help me to made me understand.
what if we choose y=±1 and x=0, is choosing 0 and ±1 for x,y going to give lowest value?
 
 
3 hours later…
10:47 AM
@user629353 It's the same argument since a,b are to be found later, thus ax+by is basically the same as ay+bx
 
 
2 hours later…
12:57 PM
Hi everyone, "local homology groups depend only on local topology provided that points are closed" - I was reading that and can see why it might not/can't? work if points are not closed, and created some examples on finite topologies for that - but I would like to see an example which is significant/practically important etc.
Do someone have a good example for that?
 
Can someone here actually give an intuitive (i.e., avoiding equations as much as possible) answer to the following question?
6
Q: What are discrete and fast Fourier transform intuitively?

mathI have done both of these in my math courses, but without understanding what they actually are intuitively. I would be very much grateful if you could give me an intuitive explanation of them.

 
1:18 PM
In one or two sentences, what is the basic idea behind discrete Fourier transform, Fourier analysis, fast Fourier transform, etc., etc.? Please, one or two sentences for each of these concepts. Each of these one-two sentences paragraph should summarise the corresponding concept.
 
2:14 PM
@KonformistLiberal The excisive triple you're using to show that statement is not necessarily excisive if points are not closed
 
@AlessandroCodenotti Yeah I know that, and as I said I created some examples even, but I was looking for a good example
 
 
2 hours later…
4:04 PM
@EmilioPisanty 'Cause I saw where it said "edited by Emilio Pisanty" and misread it as "posted by Emilio Pisanty" by accident
Sorry
 
I think I just need to use the CRT but still . . .
in Group Theory, 16 mins ago, by Shaun
in In the search of a question, 1 min ago, by Shaun
I'm looking for a solution to Exercise 4.62 of Gallian's Contemporary Abstract Algebra.
Basically: Find the number of generators of $U(49)$ without looking at the elements.
Okay, I think I understand it now.
 
$U(49)$ being $(\Bbb Z/(49))^\times$?
 
Hello!!

I want to calculate the solution $p\in \mathbb{P}^4$ of the interpolation exercise $$p(0)=2 , \ p'(0)=3, \ p''(0)=1 \\ p(1)=2, \ p'(1)=0$$

For that we have to write a general polynomial of degree 4, $p(x)=ax^4+bx^3+cx^2+dx+e$ and then with the given coditions we have to solve a linear system. Is that correct or am I supposed to do something else?
 
@ÍgjøgnumMeg Yes, and I think the answer is $\varphi(42)$ since we're given that $U(49)$ is cyclic of order $\varphi(49)=42$.
Here $\varphi$ is Euler's totient function.
. . . of course.
 
4:20 PM
I have been stuck with an eigen boundary value problem. Have already asked two variants of it on MSE. I keep getting trivial solutions to my found EVs. It would be really helpful if anyone could have a look into it
1
Q: Eigenvalues keep giving trivial solutions everytime.

Indrasis MitraI am trying to find the eigenvalues of this Eigen BVP. $\mu$ is the eigenvalue parameter $$ \lambda_h F''' - 2 \lambda_h \beta_h F'' + \left( (\lambda_h \beta_h - 1) \beta_h - \mu \right) F' + \beta_h^2 F = 0 $$ wit BC(s) $F(0)=0,\frac{F''(0)}{F'(0)}=\beta_h,\frac{F''(1)}{F'(1)}=\beta_h$ For $\...

 
@Shaun right, $49 = 7^2$ so $\varphi(49) = \varphi(7^2) = 7 \cdot 6$
woops, you mean $\varphi(49)$, not $\varphi(42)$
(which is 42, probably a typo)
@Shaun can you explain why it has order $\varphi(49)$ though?
 
@ÍgjøgnumMeg The order of $U(49)$ is $\varphi(49)=42$, since, by definition of the group of units and of $\varphi$, that is how many elements of the group there are; however, there is $\varphi(42)(=12?)$ generators of $U(49)$ because we are given that the group is cyclic.
I hope that's right. I could be wrong.
Tag me if you'd like to correct/verify me, please :)
 
@Shaun woops sorry, I didn't read your question properly
yes the units of a group with the form $\Bbb Z/(p^k)$ is a cyclic group, so you are right
 
5:04 PM
that should say rin
g
but w/e
 
5:53 PM
@ÍgjøgnumMeg Exercise 4.66 ibid. states that $U(2^n)$ is not cyclic for $n\ge 3$.
 
That has to do with $2$ not being an odd prime
 
@Shaun Right, it has an extra factor of order $2$, generated by $-1$ (the big factor being generated by $5$).
 
for example if you look at $n = 3$ and look at the elements you have $1, 3, 5, 7$ coprime to $8$ but $1^2 \equiv 3^2 \equiv 5^2 \equiv 7^2 \equiv 1 \bmod 8$ so you have $(\Bbb Z/(2^3))^\times \cong \Bbb Z/(2) \times \Bbb Z/(2)$
 
Thank you :)
 
 
2 hours later…
7:40 PM
Is oeis.org/A000569 the number of (1,-1), (1,0), (1,1) lattice paths that never drop below zero?
 
someone help me on measure theory
 
Oops nvm
 
[Now grading: The rep theory problem.]
 
if I have $\mu((A\cup B)^c)<t$ can say that $\mu((A\cup B))>t $ ?
 
@Vrouvrou where $\mu$ is a measure?
 
7:53 PM
yes
 
then no
that would be really weird
 
Sanity check: $\Bbb Z[X]/(p, (X+a)(X+b)) \cong \Bbb F_p \times \Bbb F_p$ as rings?
or is that ludicrous
 
As long as $a\neq b$ it seems to make sense to me, but don't trust me on algebra lol
 
correct, if $a\neq b$.
 
Right
 
8:03 PM
otherwise you get a local ring
 
Ah, I was thinking of $\Bbb F_p[\sqrt{-a}]$ or something like that
 
so if I have, for instance, $\Bbb Z[X]/(5, (X + 1)(X^2 + 2))$ then this is just $\Bbb F_5 \times \Bbb F_{25}$ (-2 is not a QR mod 5)
just thinking about somebody's question on the main site
 
@TobiasKildetoft please look to the question :math.stackexchange.com/questions/3079756/…
 
counting idempotent elements in that ring
@Alessandro ramification in number fields works by analysing such quotients
 
Yeah I did some similar stuff in ANT last year
 
8:08 PM
I remember feeling like I'd found some cheat codes
 
I kinda regret doing algebraic geometry instead of class field theory this semester, I think I might have enjoyed it more
 
Alggeo is cool, but turns out it's also hard
 
@ÍgjøgnumMeg Actually, you need $a$ and $b$ to not be equal mod $p$.
 
right because when you pass to $\Bbb F_p[T]$ you get $a = b$ so you're quotienting by $(X + a)^2$ again
 
8:13 PM
Oh, right, I had quotiented by $p$ automatically before even worrying about $a$ and $b$ earlier
 
So many students who claim that $1+2+2$ is the only way to write $5$ as the sum of three positive integers.
 
They must be a special kind of ultrafinitist refuting 3
 
All positive integers can be partitioned in one and only one way as a sum of exactly three positive integers
 
8:19 PM
presumably just wishful thinking on their part, because it would make them done with that part without having to argue further.
also several that claim that $1^2 + 2^2 + 2^2 = 10$.
 
they are right modulo $1$
 
@TobiasKildetoft Those are the physics inclined students
 
hahahaha
 
I know it will seem trivial, but why do we care about joint distributions? Of course, we may care about the joint distribution of two or more variables, because we may be looking for answers to questions like "What is the probability that today will rain and my friend will visit me", etc.
Anyway, I am not really asking a question which I've just answered. I didn't finish it. Why do we particularly care about joint distributions in the context of causality, if any one here knows what I am talking about?
 
So the subcategory of $\mathsf{Sch}/X$ of schemes $\pi:Y\to X$ where $\pi$ is affine is equivalent to the category of quasicoherent $\mathcal O_X$-algebras, is the full slice category $\mathsf{Sch}/X$ equivalent to a nice category of algebraic objects?
 
8:36 PM
A commutative Artinian ring has $2^m$ idempotents, where $m$ is the number of maximal ideals (finite in such a ring) of the ring
lol
cool fact?
 
8:56 PM
finally someone who managed to do the last part correctly.
 
9:21 PM
@TobiasKildetoft what was the question? (asking for a friend)
@ÍgjøgnumMeg isn't that just another way of saying that it splits as the product of $m$ non-zero rings?
the idempotent of any ring A corresponds to clopen subsets of Spec(A) anyway
@ÍgjøgnumMeg the trick is to "pull the $p$ out", i.e. 3rd isomorphism, i.e. $\Bbb Z[X]/(p,(X+a)(X+b)) = \Bbb F_p[X]/((X+\overline a)(X+\overline b))$ and $\Bbb Z[X]/(5,(X+1)(X^2+2)) = \Bbb F_5[X]/((X+1)(X^2+2)) = \Bbb F_5[X]/(X+1) \times \Bbb F_5[X]/(X^2+2)$
 
@LeakyNun (this is part 4, but I will just let you work out the entire thing with no helper-questions). Let $G = D_5\leq S_5$ and let $V$ be the permutation representation coming from the action on $\{1,2,3,4,5\}$. Decompose this as a sum of irreducibles and find the characters of the summands.
 
great, I'll try
 
(as a rep for $G$, not for $S_5$ of course).
 
I should be able to work this out myself if I want to continue studying modular rep
 
just 28 more exams to grade now (well, to grade the last problem in, as I do one problem at a time).
 
9:26 PM
you now have 29 :P
 
I will let chat grade yours
 
@Leaky well the proof I saw splits $(0)$ into $$\bigcap_{i = 1}^m \mathfrak{m}_i^k$$ so that $R \cong \prod_{i = 1}^m R/\mathfrak{m}_i^k$ and uses that each of the factors there is a local ring (wherein the only idempotents are $0$ and $1$)
 
Is there a special name for the homomorphism $\phi : G^* \rightarrow G$ such that $\phi(\varepsilon) = e_G$ and $\phi(as) = a \phi(s)$?
 
I don't know how to prove that such a decomposition for $(0)$ exists tho
 
@user76284 what does the star mean?
 
9:37 PM
Kleene star.
i.e. the set of strings of elements of G.
It's just multiplying the elements of a string together.
 
probably best to say that explicitly henceforth, most people will parse some sort of dual.
 
okay
 
it's the counit isn't it
 
as far as I can tell this is better known as "$G^\infty$", aka the union (direct limit) of $G^n$ as $n \to \infty$
 
isn't it the free group generated by $G$
oh, just free monoid
 
9:40 PM
nope, misparsed the question. leaky nun is correct.
seems reasonable to just say "the canonical map".
 
Yeah, free monoid. I'm using this map to talk about words and I want to keep a clean distinction between a word and the group element it evaluates to.
 
but then it should be $\phi(as)=\phi(a)\phi(s)$?
 
a here is an element of G and phi(a) for elements of G seems like it should evaluate to a.
 
or maybe you mean $a \in G$
 
that gives you the desired relation then for arbitrary a ofc
 
9:41 PM
Yeah, I meant for $as$ to mean the first element is $a$ and the rest is a string $s$.
 
I would just say "The canonical map" or "the forgetful homomorphism", where it "forgets" that these are words and just multiplies them together
 
9:58 PM
Hi @KarlKronenfeld.
 
not much
you?
 
busy but good
 
I never asked last year how things went
so I'm not sure what you're up to now
 
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