What Dirichlet doesn't tell... - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T09:11:23Z http://mathoverflow.net/feeds/question/72180 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72180/what-dirichlet-doesnt-tell What Dirichlet doesn't tell... Xandi Tuni 2011-08-05T14:03:01Z 2011-08-25T02:38:29Z <p>Let $n>1$ be an integer, and let us consider the set $P(n)$ of all prime numbers $p$ such that $p$ is not congruent to $1$ modulo $n$. Dirichlet's Density Theorem tells us that $P(n)$ has a natural density, equal to $$1-\varphi(n)^{-1}$$ where $\varphi(n) = |(\mathbb Z /n)^\ast|$ is Euler's totient.</p> <p>From the Frobenian point of view, saying that $p$ is congruent to $1$ modulo $n$ is to say that the ideal $(p)$ splits completely in the cyclotomic field $\mathbb Q(\zeta_n)$. </p> <p>From Chebotarev's point of view, saying that $p$ is congruent to $1$ modulo $n$ is to say that the Frobenius element over $p$ in $\operatorname{Gal}(\mathbb Q(\zeta_n)|\mathbb Q) \simeq (\mathbb Z /n)^\ast$ is the identity.</p> <p>So far so good, now let us consider the set $P$ of all prime numbers $p$ which are not congruent to $1$ modulo $n^2$ for any $n>1$, that is $$P := \bigcap_{n>1}P(n^2) = \bigcap_{\ell\mathrm{ prime}}P(\ell^2)$$ Supposing that "the events $P(\ell^2)$ are uncorrelated" for different $\ell$'s, we can phantasise about the density of $P$, hoping it might be (at least up to a rational factor, I don't vouch for it) $$\operatorname{dens}(P) = \prod_\ell 1-\frac{1}{\ell(\ell-1)} \quad = 0.37395581361920228805...$$ a number called <em>Artin's constant</em> (it appears in Artins primitive root conjecture, which is similar in nature). The question whether $P$, or similarly constructed sets of primes, have a density and whether it is the expected one goes far beyond the density theorems of Dirichlet, Frobenius and Chebotarev. The corresponding Galois extension would be the maximal cyclotomic extension of $\mathbb Q$, which is ramified everywhere. </p> <blockquote> <p>Can you name this problem? Have you seen it before? Where? </p> </blockquote> <p>Hooley (1967) has shown that Artins primitive root conjecture follows from GRH. In principle, the problem of determining the density of $P$ should be simpler. </p> <blockquote> <p>Under GRH, is it true that the density of $P$ exists and is equal to Artin's constant?</p> </blockquote> http://mathoverflow.net/questions/72180/what-dirichlet-doesnt-tell/72181#72181 Answer by Gjergji Zaimi for What Dirichlet doesn't tell... Gjergji Zaimi 2011-08-05T14:19:25Z 2011-08-05T14:19:25Z <p>The specific density result you quote is a result of Mirsky, see</p> <blockquote> <p>L. Mirsky, "The number of representations of an integer as the sume of a prime and a k-free integer", Amer. Math. Monthly 56 (1949)</p> </blockquote> <p>There have been several generalizations, for the direction on replacing squares with higher powers see "Values of the Euler function free of k-th powers" by W.D. Banks and F. Pappalardi. For the result on primes not congruent to $\frac{a}{b}\pmod{n^2}$ see "Arithmetic progressions, prime numbers, and squarefree integers" by S. Clary and J. Fabrykowski.</p> http://mathoverflow.net/questions/72180/what-dirichlet-doesnt-tell/72191#72191 Answer by Anonymous for What Dirichlet doesn't tell... Anonymous 2011-08-05T16:34:41Z 2011-08-05T16:34:41Z <p>Let me sketch a proof. If you <strong>fix</strong> a bound $z$, then the events $P(\ell^2)$ for different $\ell \leq z$ <strong>are</strong> uncorrelated; this is just a consequence of the prime number theorem for progressions and the multiplicativity of Euler's function. This reduces the problem to "understanding the tails"; in other words, we have to show that as $z\to\infty$, the relative upper density of the primes divisible by $\ell^2$ for some $\ell > z$ tends to zero.</p> <p>Consider the primes $p \leq x$. By the Brun--Titchmarsh inequality, the number of such $p$ for which $p-1$ is divisible by $\ell^2$ for some prime $\ell$ with $z &lt; \ell \leq x^{1/4}$ (say) is $$\ll \sum_{z&lt; \ell \leq x^{1/4}} \frac{x}{\phi(\ell^2)\log{(x/\ell^2)}} \ll \pi(x) \sum_{z > \ell} \frac{1}{\ell^2} \ll \frac{\pi(z)}{x}.$$ Also, the number of $p$'s with $p-1$ divisible by $\ell^2$ for some $\ell > x^{1/4}$ can be estimated trivially: We just count how many $n \leq x$ are divisible by some $\ell^2$ with $\ell > x^{1/4}$, which is clearly at most $\sum_{\ell > x^{1/4}} \lfloor x/\ell^2\rfloor \ll x^{3/4}$. This is negligible for us. Hence, the relative upper density in the previous paragraph is $\ll 1/z$. So it does indeed tend to zero as $z\to\infty$.</p>