nt.number theory - Sum of reciprocals of primes modulo which a polynomial has a root

Charles is completely right that this follows from Frobenius' theorem. Since you don't like Galois theory, here is a proof which does not explicitly mention Galois theory. (But it is hiding just out of sight.)

We may assume that $f$ is irreducible as, if $g$ divides $f$, then the set of primes for which $f$ has a root contains the set for which $g$ does.

Let $K$ be the field $mathbb{Q}[x]/f(x)$. Let $R$ be the ring of integers of $K$, and let $S=mathbb{Z}[x]/f(x)$. Note that $S$ is a finite index sublattice of $R$ and, if $p$ is a prime which does not divide $|R/S|$, then $R/p cong S/p$. Also, for any prime $p$ which does not divide the discriminant of $f$, the polynomial $f$ factors into distinct factors in $mathbb{F}_p[x]$.

Thus, if $p$ is large enough to not divide either $|R/S|$ or the discriminant of $f$, then $R/p cong S/p cong mathbb{F}_p[x]/f(x) cong bigoplus mathbb{F}_p[x]/(f_i(x))$ where $f_i$ are the irreducible factors of $f$ mod $p$. So, for such a prime $p$, prime ideals of $R$ which contain $(p)$ are in bijection with irreducible factors of $f$ mod $p$, and the norm of such a prime is $p^{deg f_i}$.

So, if $f$ has a root modulo $p$, then
$$frac{1}{p} leq sum_{pi supseteq (p), pi mbox{prime}} frac{1}{N(pi)} leq frac{deg f}{p}$$
and, if $f$ does not have a root modulo $p$, then
$$sum_{pi supseteq (p), pi mbox{prime}} frac{1}{N(pi)} leq frac{(deg f)/2}{p^2}$$

We want to show that
$$sum_{p: exists pi mbox{a prime of} R mbox{with} N(pi)=p} frac{1}{p}$$
diverges. By the above inequalities, it is equivalent to show that
$$sum_{pi subset R, pi mbox{prime}} frac{1}{N(pi)}$$
diverges. (Note that the finitely many primes which divide $|R/S|$ or the discriminant of $f$ cannot change whether or not the sum converges.)

Now, we have unique factorization into prime ideals for $R$, so
$$sum_{I subseteq R} frac{1}{N(I)^s} = prod_{pi subset R, pi mbox{prime}} left( 1 - frac{1}{N(pi)^s} right)^{-1}.$$

The left hand side is the $zeta$ function of $K$. By the class number formula (see most books on algebraic number theory), $zeta_K(s) = C/(s-1) + O(1)$ for some positive constant $C$, as $s to 1^{+}$. So
$$log zeta_K(s) = log frac{1}{s-1} + O(1) = sum log left( frac{1}{1-N(pi)^{-s}} right) = sum frac{1}{N(pi)^s} + O(1/N(pi)^{2s}).$$
We deduce that
$$sum frac{1}{N(pi)^s} = log frac{1}{s-1} + O(1)$$
so
$$sum frac{1}{N(pi)}$$
diverges.

co.combinatorics - Maximum differences in sorted vectors of naturals

This question is related to one I asked previously. This is probably a little harder. I had a crack at it today, but have become stuck. I suspect the result is buried in the order statistics literature somewhere, and perhaps somebody is familiar with it. That, or Peter might insta-solve again :).

Given a vector $s$ of integers let $d(s)$ be the maximum difference between any two integers in $s$ when sorted in ascending order. That is, if we sort $s$ in ascending order to obtain $v$, then
$$d(s) = max_{i} (v_{i+1} - v_i).$$

For $s$ a vector of length $m$ from $lbrace 1,2,dots,nrbrace^m$ we must have $0 leq d(s) < n$.

Given $0 leq k < n$, how may such vectors have $d(s) = k$ ?

Again, I'm more interested in the case where $n$ is much larger than $m$ and if reasonable bounds can be found for $d(s)$, then this would be useful too.

Note: If $N_k$ is the answer for $k$. Then you should have $n^m = sum_{k=0}^{n-1}N_k$

ct.category theory - Internal hom of sheaves

Well, here is a partial answer. The category of abelian group-valued sheaves is not a topos, the category of set-valued sheaves is. And I think you should look at set-valued presheaves, at least $hom(-,U)$ is one:

When the site comes from a topological space, you can see as follows that your two definitions coincide: When you insert into the $hom$ in your second expression an open set $V$, you either get $hom(V,U)=$the one-element-set containing only the inclusion, if $V subseteq U$, or $hom(V,U)=$empty set, if not, so taking a product with the set-valued presheaf $hom(-,U)$ either leaves X as it is or "deletes" it (i.e. transforms it into the (presheaf with value the) empty set).

So a natural transformation in $hom_{PreShv}(X times hom(-,U), Y)$ is given just by its components for all $V subseteq U$ and for other $V$ it extends to the unique map from the initial object into Y. This is the same as a natural transformation of restricted presheaves.

Edit: It now occurred to me that you were probably actually speaking of the sheaves of group homomorphisms, not just any natural transformations. You get that, when you apply the "free group"-functor to the set-valued sheaf "hom(-,U)" and take tensor product instead of product. The free group over the empty set ist the trivial group, thus the initial object in groups and tensoring with it trivializes $X$. Tensoring with the free group over one element doesn't change anything, so the same things happen as in the set case...

For more general sites, the expression $X| _U$ probably means that you look at $X$ as a functor defined on the slice category given by maps on your site into its object $U$. But you can also see $X$ as a sheaf not just on the site, but on all of its ambient sheaf category (as a hom-functor, and where you give to topos an appropriate topology - you sometimes do this in topos theory). Then looking at restricted sheaves is the same as passing to the slice category $Shv / hom(-,U)$, and $X times hom(-,U)$ is the image of $X$ under the canonical functor $Shv rightarrow Shv/hom(-,U)$ (which is given by taking product with $hom(-,U)$), so it actually $is$ $X| _U$. This still doesn't make it clear to me why the two functors in question coincide, but I guess it may be a way to look at it to find it out...

integration - Inverse of a function defined by an integral

You can always get a (non-linear) ordinary differential equation for $f^{-1}$. It is easy to see that $f$ satisfies a 2nd order linear ODE with polynomial coefficients with no order 0 term [the first order ODE has non-polynomial coefficients, so harder to work with]. From there, it is also mechanical to get an ODE for $f^{-1}$ by interchanging the roles of $f$ and $w$.

But since your original function is (in general, depending on the path of integration) a MeijerG function, very few of these have closed-form inverses. As Piero mentions in the comments, the trigonometric and Elliptic functions are some of the few cases where this 'works'.

It also depends on what you are trying to do 'next' with these functions. If you are looking at numerical evaluation, then there are closed-forms for the Lagrange inverse, some of which translate to closed-forms for the Hermite-Pade approximants, from which efficient approximations can be derived. [J.M.'s method works in general, here it turns out that this can be pushed in closed-form further than usual].

For any given case, there are useful tools in Maple (and Mathematica) to help you carry these computations out. Probably in other CASes as well, but I don't know.

math philosophy - Does the Golden Ratio Apply to Timing as Well?

My guess this is not an appropriate question for MO, but being weekend and all...

I have heard of one approach to memorization which consists in repeating something in time intervals corresponding to the Fibonacci sequence. I did not find a reference, so take this with a grain of salt, though :)

nt.number theory - Why Weil group and not Absolute Galois group?

The Weil group appears for several reasons.

Firstly: if $K$ is a non-archimedean local field with residue field $k$,
the local reciprocity law induces an embedding
$K^{times} hookrightarrow G_K^{ab}.$ The image consists of all
elements in $G_k$ whose image is an integral power of Frobenius.
This is the abelianized Weil group; it just appears naturally.

Secondly: suppose that $K$ is a global field of positive characteristic,
i.e. the function field of a curve over a finite field $k$. Then the
global reciprocity map identifies the idele class group of $K$ with
a subgroup of $G_K^{ab}$ consisting of elements which act on $k$
by integral powers of Frobenius. So again, it is the abelianized Weil group
that appears.

Thirdly: suppose that $E$ is an elliptic curve over a quadratic imaginary
field $K$
with complex multipliction by $mathcal O$, the ring of integers in $K$. (Thus I am implicity fixing $K$ to
have class number one, but this is not so important for what I am going to say next.)
If $ell$ is a prime, then the $ell$-adic Tate module is then free of rank one over
$mathcal O_{ell}$ (the $ell$-adic completion of $mathcal O$), and the $G_K$-action
on this Tate module induces a character $psi_{ell}:G_K^{ab} rightarrow mathcal O_{ell}^{times}$.

There is a sense in which the various $psi_{ell}$ are indepenent of $ell$,
but what is that sense?

Well, suppose that $wp$ is a prime of $K$, not dividing $ell$ and at which
$E$ has good reduction. Then the value of $psi_{ell}$ on $Frob_{wp}$ is indepenent
of $ell$, in the sense that its value is an element of $mathcal O$, and this value
is independent of $ell$.
More generally, provided that $wp$ is prime to $ell$,
the restriction of $psi_{ell}$ to the local Weil group at $wp$ is independent of $ell$
(in the sense that the value at a lift of Frobenius will be an algebraic integer that
is independent of $ell$, and its restriction to inertia at $wp$ will be a finite image
representation, hence defined over algebraic integers, which again is then independent
of $ell$).

Note that independence of $ell$ doesn't make sense for $psi_{ell}$
on the full local Galois group at $wp$, since on this whole group it will
certainly take values that are not algebraic, but rather just some $ell$-adic
integers, which can't be compared with one another as $ell$ changes.

Now there is also a sense in which the $psi_ell$, as global Galois characters, are independent of $ell$. Indeed, we can glue together the
various local Weil group representations to get a representation $psi$ of the global Weil
group $W_K$. Since it is abelian, this will just be an idele class character $psi$,
or what is also called a Hecke character or Grossencharacter. It will take values
in complex numbers. (At the finite places it even takes algebraic number values, but
when we organize things properly at the infinite places, we are forced to think of it
as complex valued.)

Note that $psi$ won't factor through the connected component group, i.e. it won't be
a character of $G_K^{ab}$. It is not a Galois character, but a Weil group character.
It stores in one object the information contained in a whole collection of
$ell$-adic Galois characters, and gives a precise sense to the idea that these various $ell$-adic characters are independent of $ell$.

This is an important general role of Weil groups.

Fourthly: The Hecke character $psi$ above will be an algebraic Hecke character, i.e.
at the infinite places, it will involve raising to integral powers. But we can also
raise real numbers to an arbitrary complex power $s$, and so there are Hecke characters
that do not come from the preceding construction (or ones like it); in other words,
there are non-algebraic, or non-motivic, Hecke characters. But they are abelian characters
of the global Weil group, and they have a meaning; the variable $s$ to which we can raise
real numbers is the same variable $s$ as appears in $zeta$- or $L$-functions.

In summary: Because Weil groups are "less completed", or "less profinite", than Galois
groups, they play an important role in describing how a system of $ell$-adic representations can be independent of $ell$. Also, they allow one to describe
phenomena which are automorphic, but not motivic (i.e. which correspond to non-integral
values of the $L$-function variable $s$). (They don't describe all automorphic phenomena,
though --- one would need the entire Langlands group for that.)

arithmetic geometry - Decomposition of Tate-Shafarevich groups in field extensions

Fix a prime p which doesn't divide the degree of K over ${mathbb Q}$, and let ${mathcal O}$ denote the ring of integers of ${mathbb Q}_p(chi)$ i.e. an extension of ${mathbb Q}_p$ containing the values of $chi$. Then the group algebra ${mathcal O}[G]$ decomposes into a direct sum of 1-dimensional pieces over ${mathcal O}$, one for each power of $chi$.

Then $Sha(E/K)[p^infty] otimes {mathcal O}$ being an ${mathcal O}[G]$-module inherits such a decomposition. Concretely, the $chi^i$-component of $Sha(E/K)[p^infty] otimes {mathcal O}$ is the subset where $G$ acts by $chi^i$.

This $chi^i$-component is then a reasonable candidate to compare to the $p$-adic valuation of the algebraic part of $L(E,chi^i,1)$.

Some further comments:

-Note that extending scalars to ${mathcal O}$ increases the size of the modules so this has to be taken into account.

-The component corresponding to the trivial character is the invariants under $G$, and when $G$ has size prime-to-p this is simply $Sha(E/{mathbb Q})[p^infty] otimes {mathcal O}$ (which is good).

-To make a precise relationship between the $L$-value and Sha, you need to take into account the other terms in BSD. Namely:

*The torsion-term should work out exactly as above (decomposing into $chi$-components).

*The periods have to be considered (which was ignored above in my vague phrase "the algebraic part of").

*The Tamagawa numbers give me pause -- possibly there is an analogous $chi$-decomposition, but I don't see it now.

*Lastly, if K is ramified over ${mathbb Q}$ then the discriminant of K appears in the BSD quotient (in the denominator which increases the size of Sha). To handle this, I imagine what should be done is that rather then considering the L-value alone, consider the L-value times the Gauss sum of the character. (By the conductor-discriminant formula this should give exactly the extra powers of p needed.)