JSchlather

@JasperLoy I'll probably just stick with JSchlather I've seen another Jacob here and there.

Otherwise in a lot of situations I think the absolute Galois group is poorly understood.

unless I can reduce to the case $k(t)$ somehow.

Well, I think in general Galois theory won't be too useful

right

That is neat. It makes sense since profinite groups are groups that are projective limits of their finite subgroups.

Okay, yeah. I couldn't think of an object off the top of my head where that property wouldn't hold.

@anon I haven't studied it much and I needed it to have a property which I decided it had. Namely that the profinite free group on $\kappa$ generators contains the profinite free group on $\mu$ generators where $\mu,\kappa$ are infinite (actually uncountable) cardinals such that $\mu \leq \kappa$.

yea

@AlexanderGruber Do you know anything about the free profinite group?

Are there are any set theorists around?

@Sanchez I posted an answer after deciding it was true.

@Sanchez Yeah, I was trying to do something simple like $(x-\sqrt[p]{t})^p$. But I couldn't get it to work.

@Sanchez Did you have a counterexample in mind for math.stackexchange.com/questions/254455/… this question? I was reading it again and it seems like if you assume the $u_i$ are the distinct roots of $f$ it may be true.

@BenjaLim What are you talking about?

Yes, eric but what is 4 in the left component?

I meant subgroup

Oh sorry

It's the ideal generated by (2,2).

But how can you say it exactly in words when the english language is so imprecise ;)

@JacobBlack Yea, no more pings.

Yeah that works

yeah

Oh soryr

@BenjaLim arctan(x pi/2)

Yeah, I guess it is. Who knows why Evans added it. Thanks.

Yes

I agree. So you don't see any need for the sqrt(n) term either?

@BenjaLim Also, you clearly need to expand your musical horizons ;).

So I don't understand where the $\sqrt{n}$ term is used at all.

Then putting this in to the previous bound we have that $|Du(x_0)| \leq \frac{\sqrt{n}C_1\alpha(n)}{r} ||u||_{L^\infty (\mathbb{R}^n)}$

So we can bound the L^1 norm by $\alpha(n) r^n ||u||_{L^\infty(\mathbb{R}^n)}$

(sorry)

yes

so he says that |Du(x_0)| \leq sqrt(n) * other bound

But he introduces a sqrt(n) term into the previous bound which I don't understand the need for

To prove that a harmonic function which is bounded is constant he uses this bound and then relates it to the l^infty norm of u.

and alpha(n) is the dimensionality constant

where C_1=2^(n+1)n/alpha(n)

For a harmonic function u in R^n we have that $|Du(x_0)| \leq (C_1/r^{n+1}) ||u||_{L^1(B(x_0,r))}$

So essentially the important parts are as follows

Yeah

I could link you to an illegal copy of Evan's.

He proves it by first giving a bound for derivatives of a harmonic function then uses this to deduce liouville's theorem, but he throws in a $sqrt{n}$ that I can't figure out.

@robjohn I'm having an issue with the proof of Liouville's theorem for harmonic functions in Evan's PDE book.

No, refer to last.fm/user/liberalkid

The rapper?

There's a proof in Evans where he throws in a $\sqrt{n}$ for no reason

@BenjaLim :(

Anyone knowledgeable at PDE on hand?