My profile pic is a section of my desktop background

Shelah proved that, Shelah also showed there there are uncountable groups where very proper subgroup is countable, although I don't think that used forcing

Another surprising statement independent of ZFC is the projective dimension of $\Bbb R(x,y,z)$ as a $\Bbb R[x,y,z]$ module, it is $2$ iff CH holds and $3$ otherwise

@Ultradark I generated my desktop background using python. I also made a google chrome theme for it to use!

@Spencer1O1 what is project Euler? I'm going to look at the magic squares problem

That is weird

@PaulPlummer It's faster to list the results Shelah didn't prove :P

Haha

@Ultradark It is a series of math problems that you solve with computer programming.

also spencer, I went a little low-tech with my profile picture

I made it on desmos.com

@Ultradark projecteuler.net

@PaulPlummer See the answers here

Actually they have $\Bbb C$ in the answer there, but it should be the same with $\Bbb R$

@Spencer1O1 I'm trying to make a plot of the irreducible and reducible polynomials of the form $x^n+(1-x)^r$

where $x,r$ are positive natural numbers

and $x$ is between $0$ and $1$

irreducible polynomials are sort of like primes in that they cannot be factored further

@Ultradark I hate irreducible polynomials because you can't easily find the zeroes.

But I guess they don't come up very much anyway

@Ultradark why are you plotting them for $x$ in that range?

So far I've found that $x^n+(1-x)^n$ is irreducible for $n=2,3,4,5,7,8,11,13$

Cool

@TobiasKildetoft because I'm representing each polynomial as a point in the unit square. and coloring the irreducible and reducible polynomials different colors

Related to a question I asked on MO a whileI don't know if it is open): Assuming continuum hypothesis is false, can a countable group have exactly $\aleph_1$ non-isomophic quotient groups? It turns out for finitely generated groups that it is either countable or continuum. @AlessandroCodenotti

so for example I'm taking $x^3=(1-x)^3$ and plotting it at a certain height

of course $x=1/2$ is the solution

A polynomial can be reducible and still have no roots.

Very important lesson in beginning algebra courses.

thanks I'll keep that in mind

So for the second part of this, can I just say "a plane"?

Someone suggested I try coming up with a better counterexample but I can't think of one

YCor's argument in the answer to your question is very neat @Paul

And that's an interesting question

I guess it is pretty standard descriptive set theory stuff

DogAteMy: Parabolic cylinder.

I always wanted to properly learn descriptive set theory but never did

@TedShifrin Rotating $y=x^2$ across the y-axis? That has no lines (and also has $K>0$)

No, cylinder ....

I've been reading some cool stuff concerning AD lately though

Axiom of determinacy? In undergrad I knew someone who worked quite a bit with that. It is a cool axiom

@TedShifrin What is that? Just the graph of $y=x^2$ in 3D?

Yup.

That's isometric to a plane

That is flat, too, of course.

as a Riemannian surface

Yeah, it came up in the answer to this question I asked on MSE so I've been looking at the references suggested by Caicedo

But it has nonzero mean curvature.

Exercise: At any point of a line in a surface you must have $K\le 0$.

I used representation theory to solve this lol

10

$K$ here is the sectional curvature, which for 2-manifolds is the Gaussian curvature

I'm a bit aware of that.

In higher dimensions you can concoct more interesting examples.

@LeakyNun I used that as a motivating example last time I lectured in algebra

@LeakyNun Does this involve $e$

(you can also easily find the variance)

I was also very fond of Burnside's Theorem when I taught group theory.

@TobiasKildetoft nice!

@AkivaWeinberger no

@AkivaWeinberger the problem is very easy from straight probability

@AkivaWeinberger That would be if you wanted the probability that nobody got their own hat

$1/e$ shows up as the probability of no letter in the right envelope, with infinitely many :P

but he used representation theory

Ah right

Well at least I'm not coming from nowhere

You can also approach such questions with Burnside.

@ÉricoMeloSilva for some odd reason I never thought of your solution

Is it just "expected number of times A1 gets his hat" plus "expected number of times A2 get his hat" etc

'cause expectation is linear even if stuff isn't independent

yes

Ohh, and of course you can also (in case you wanted to) find the expected value of the square of the number of people who got their own hat.

our guests are now prisoners

@TobiasKildetoft I thought you already said that

Wait now I'm confused

I'm getting 1 as the answer

that is the answer

1/n

@LeakyNun No, that is not the same as the variance

Plus 10

@ÉricoMeloSilva Oh, OK. That's surprisingly low

@AlessandroCodenotti Caicedo was his advisor!

@TobiasKildetoft yeah sure but Var(X) = E(X^2) - E(X)^2

so given E(X) they're "interconvertible"

ahh, sure

@PaulPlummer I guess set theory is a very small world :P

So for the square we need to see how much A1 and A2 depend on each other, or something?

E[A1 and A2] would be 1/(n(n-1)) I think

We don't end up with $\frac1n+\frac12$ for the square, do we?

it should still be a whole number, from rep theory

I think it's 2

@AkivaWeinberger No, we end up with $2$

my rep theory isn't that rusty :P

So I've gone wrong somewhere

Which fits with the above formula since the variance is $1$

Oh I think I see where I went wrong

Ahh got it

Right OK so it's $n\cdot\frac1n+\binom n2\cdot\frac1{\binom n2}$

which is $2$

Do we end up with the cubes being 3?

@AkivaWeinberger I was just thinking about it, and I did not see a particularly nice way to do this using rep theory, so I didn't want to do it

rep theory says at least 3... I'm also thinking about how to do it using rep theory

OK revised conjecture

2.5

well as I said, rep theory says at least 3

@AkivaWeinberger still an integer by rep theory

Hm arright never mind then

Oh wait I see where I messed up

4

Final answer, for real this time

we have two maps $V \to V \otimes_G V$ right

See I'm revising $1+\frac12+1$ to $1+\frac21+1$ so it's not a huge jump

or let's just decompose $V \otimes_G V = \Sigma^2 V \oplus \Lambda^2 V$

$4$ sounds plausible

@AkivaWeinberger for n=4 it is 5

I have a qeustion. $1$ person is at a table and he brings his friends to the table and sometimes tells them to leave. The number of the people at the table is always prime. Furthermore the person alternates between inviting and telling people to leave. what is the expected number of poeple at the table under the assumption that this person has $N$ friends.

@TobiasKildetoft I have a Sylow-theoretic proof that Q8 doesn't embed in S7, if you're interested

maybe you already know

@LeakyNun Never really thought about it

should I tell you?

@AkivaWeinberger for n=5 it is still 5

@TobiasKildetoft because after it I have a generalized problem

@LeakyNun Skipping 3?

@LeakyNun I am on my way to bed right now

@TobiasKildetoft ok then

@AkivaWeinberger for n=3 it is still 5

3, 2+1, 1+2, 1+1+1

There should be one for each of those

(forget what they're called)

Partitions? Or do those not respect order

what do you mean?

is 89 prime

oh i thought this was google

Yes, no @Ultradark

Well actually it's $(5+8i)(5-8i)$ so no

(/s)

$n=2,3,4,5,7,8,11,13,16,17,19,23,29,31,37,41,43,47,53,59,61,64,67,71,73,79,83,89‌​,97$

$|n|=30$

$32$ should be in there too

This is a sequence of numbers

I noticed twofold things about this sequence

they are all primes except $4,8,16,32$

so i have a conjecture. aside from numbers of the form $2^k$

the sequence consists of primes

The sequence comes from $x^n+(1-x)^n$

the list of $n$ above is the values for which it is irreducible

why is the polynomial irreducible for nearly all $n\in p$

where $p$ is the set of primes

irreducible over $\Bbb Q$

@AkivaWeinberger what's the smallest $|d|$ such that 89 ramifies in $\mathcal O_{\Bbb Q(\sqrt d)}$

ah that would be $d=89$ lol

for which $d$ are $89$ irreducible though

For $p$ prime $x^p+(1-x)^p$ looks like $px^{n-1}+\text{lot of stuff}-px+1$, all the coefficients of the middle terms are divisible by $p$ and you can apply Eisenstein's criterion for irreducibility after reversing the order of the coefficients (this works because a polynomial is irreducible in $\Bbb Q[x]$ iff it is irreducible in $\Bbb Q[x,x^{-1}]$ and the substitution $x\mapsto x^{-1}$ is an automorphism of the latter that reverses the order of the coefficients)

@Ultradark

The same argument should work for powers of $2$ as well but the coefficients are a bit different

@AlessandroCodenotti nice

let's find an outlier...

why?

I mean do we want to find an outlier?

Thanks I'm going to read more about Eisenstein's criteria

Problem: Let $\{f_n\}$ be a sequence of measurable functions on $E$. If for each $\epsilon > 0$, there is some $E_0 \subseteq E$ set of finite measure and $\delta > 0$ such that for each measurable $A \subseteq E$ and $n$, if $m(A \cap E_0 ) < \delta$ ,then $\int_{A} |f_n| < \epsilon$, show that $\{f_n\}$ is uniformly integrable and tight over $E$.

oh yeah of course there is no outlier...

okay so there's no outlier

because $x^5-y^5=(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)$ etc

@Ultradark because every time you think you find something, you should try as hard as possible to disprove yourself (see video below)

Let $\epsilon > 0$. Since $m((E \setminus E_0) \cap E_0) = 0 < \delta$, I believe it follows that $\int_{E \setminus E_0} |f_n| < \epsilon$ for every $n$. This shows tightness, but how do I show uniform integrability?

hmmm

interesting vid