User alastair irving - MathOverflow

Class numbers of the field generated by 2^{1/n}

2013-03-07T12:43:28Z

I have used sage to compute the class number of the number field generated by the polynomial $x^n-2$ for small $n$. Specifically, setting proof=False, (which hopefully just means that GRH is assumed), the class number is $1$ for all $n\leq 40$. This seems slightly surprising.

I have two questions:

Does anyone know if these class numbers have been computed for larger $n$? If so do we know a $n$ for which it isn't $1$?
Heuristically, how should the class numbers behave? Presumably there should be infinitely many $n$ for which its $1$, but is there a heuristic justification of why I haven't found a single $n$ where its not $1$. Obviously I've only considered very small $n$ but with other families of fields, real quadratic for example, its quite easy to find examples with class number $>1$.

Are there examples of sets containing no primes but for which both Type I and Type II information can be proven?

2012-06-22T11:10:54Z

In Harman's book "Prime Detecting Sieves" he describes a method to prove that a set contains primes if we have enough Type I and Type II information for it. As shown by Selberg's example of the set of numbers with an even number of prime factors Type I information is not sufficient to detect primes. Harman's methods require the Type II information to be given on sufficiently long intervals so that the sums we cannot give assymptotic formulae for are sufficiently small. Is it possible to prove, by giving a suitable counterexample, that if we have Type I information but only a very small amount of Type II information then we can't detect primes?

Let $A$ be the set in which we are interested, and suppose that $A\subseteq B$, the set of integers in $[x/2,x)$. Then, for a suitably small error $E$ and arbitrary bounded coefficients $a_m,b_n$ a Type I estimate is of the form $\sum_{mn\in A,m\leq M}a_m=\lambda \sum_{mn\in B,m\leq M}a_m+O(E)$ and a Type II sum is of the form $\sum_{mn\in A,x^\alpha\leq m\leq x^{\lpha+\beta}}a_mb_n=\lambda \sum_{mn\in B,x^\alpha\leq m\leq x^{\lpha+\beta}}a_mb_n+O(e).$ I am then wondering whether there are examples of sets satisfying these conditions for some $M,\alpha$ and sufficiently small $\beta$ which contain no primes. For definiteness take $M=x^{3/4}$ and $\alpha=1/4$.

Answer by Alastair Irving for Distribution of primes in small intervals

2012-03-29T21:01:18Z

A weaker question is to ask for which functions $f$ the interval $[x,x+f(x)]$ contains a prime for all sufficiently large $x$. The sharpest uncoditional result is then that $f(x)\geq x^{0.525}$ is sufficient. We are therefore a long way from being able to prove results about $f(x)=\log^c x$.

Answer by Alastair Irving for Prime factorization theory

2010-11-27T23:03:24Z

Suppose $p,q$ are prime. Thus $\omega,\Omega,\lambda$ are known constants at $p,q$ so we want to know if $\omega,\Omega(p+q)$ can be written as functions in $p,q$. To answer this you need to define what is meant by a function, as clearly $\omega(p+q)$ is a function in $p,q$.

Presumably you mean a function which is built from some class of elementary functions. Therefore, a possible approach would be to try and prove that certain classes of functions are not sufficient. For example, we cannot have $\omega(p+q)=Ap+Bq+C$, for constants $A,B,C$, as we can pick suitable combinations of primes to give a system of linear equations in $a,B,C$ with no solutions, (I haven't actually done this computation). It seems reasonable that for any degree $d$ we can find suitable pairs of primes to show that the function isn't a polynomial with degree $\leq d$, but I don't have a proof of this. Thus showing that the function isn't a polynomial might be a worthwhile first It will probably help to use that the function is symmetric in $p,q$ so if its a polynomial then its a polynomial in $p+q,pq$.

Once polynomials are dealt with, then try and extend the class of possible functions to something bigger.

Is there an elementary proof that the Mertens function is not $O(x^\theta)$ if $\theta <1/2$?

2010-11-26T15:24:15Z

The Mertens function is the partial sums of the Moebius function: $M(x)=\sum_{n\leq x}\mu(n)$ Since the zeta-function has a zero on the critical line it follows that $M(x)\ne O(x^\theta)$ for any $\theta<\frac 12$.

Does anyone know if there is an elementary proof of this statement? (By elementary I mean a proof which does not depend on complex analysis, in particular the existance of a zero of $\zeta$). even an elementary proof of $M(x)$ being unbounded would be interesting to me.

Comment by Alastair Irving

2013-03-08T11:07:19Z

Yes, I did mean the field formed by adjoining a single root of $x^n-2$, so I agree this is a very similar question to mathoverflow.net/questions/88288/ and therefore it should be closed.

Comment by Alastair Irving

2013-01-25T09:18:51Z

assuming I understand what you're asking then I think this is a very uninteresting sum, by rearranging the terms we see that $\sum kF(k)$ is simply the length of the interval $[2,p_n]$, so your result follows trivially.