User joe silverman - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T21:39:51Z http://mathoverflow.net/feeds/user/11926 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/132380/what-is-the-discriminant-divisor-of-a-surface-fibered-over-a-curve/132383#132383 Answer by Joe Silverman for What is the discriminant divisor of a surface fibered over a curve? Joe Silverman 2013-05-31T02:08:18Z 2013-05-31T02:08:18Z <p>If $X$ is a minimal elliptic surface, then (as you're probably aware) there's a standard way of defining the discriminant divisor on $C$, and in addition one has the beautiful formula of Ogg which states that for points $t\in C$, <code>$$f_t = \text{ord}_t(\mathcal{D}_{X/C}) + 1 - m_t,$$</code> where $f_t$ is the valuation of the conductor (which in this case is 0,1,2), and $m_t$ is the number of components of the fiber $X_t$. Then you get the full discriminant as the sum <code>$$\mathcal{D}_{X/C} = \sum_{t\in C} \text{ord}_t(\mathcal{D}_{X/C})(t).$$</code></p> <p>Ogg's local formula is valid for elliptic surfaces over $\text{Spec}(R)$ for a DVR, and he proved the formula for all residue characteristics except in the case that $R$ has characteristic $0$ and its residue field $R/\mathfrak{p}$ has characteristic 2. (The hard cases are characteristic 2 and 3.)</p> <p>A number of years later, in a beautiful paper, T. Saito gave a new proof that completed the elliptic surface case and generalized it to arbitrary genus fibrations. His article is</p> <p>T. Saito, Conductor, discriminant, and the Noether formula of arithmetic surfaces, <em>Duke Math J</em> <strong>57(1)</strong> (1988), 151-173.</p> <p>Saito's article will have a formula for the discriminant of your $X\to C$ with genus $g$ generically smooth fibers, computed locally above each point of $C$.</p> http://mathoverflow.net/questions/132191/proving-that-every-term-of-the-sequence-is-an-integer/132231#132231 Answer by Joe Silverman for Proving that every term of the sequence is an integer Joe Silverman 2013-05-29T15:32:53Z 2013-05-29T15:32:53Z <p>This looks like it might be a "rank two elliptic divisibility sequence", i.e., let $E$ be an elliptic curve, let $P$ and $Q$ be independent points in $E(\mathbb{Q})$, and write $$x(mP+nQ) = A(m,n)/D(m,n)^2.$$ Then the doubly-indexed sequence $D(m,n)$ is a rank two elliptic divisibility sequence, or as they were named by Kate Stange, an <em>elliptic divisibility net</em>. Kate made a detailed study of these sequences in her thesis (<em>Algebra and Number Theory</em>, 5-2 (2011), 197-229; it's also available on the ArXiv at <a href="http://arxiv.org/abs/0710.1316" rel="nofollow">http://arxiv.org/abs/0710.1316</a>), including a description of the recurrence relation(s) that they satisfy. </p> <p>Anyway, I don't know for sure if your rank two sequence fits into Kate's elliptic net framework, but it's quite possible that it does.</p> <p>(I'm actually cheating a little bit here, one really needs to use division polynomials instead of writing $x(mP+nQ)$ as a fraction in case there's cancellation between the numerator and the denominator. But this will only affect primes of bad reduction.) </p> http://mathoverflow.net/questions/131996/dynamics-in-one-matrix-variable/132046#132046 Answer by Joe Silverman for Dynamics in one matrix variable Joe Silverman 2013-05-27T20:24:03Z 2013-05-27T20:24:03Z <p>A few elementary (and probably not very useful) thoughts:</p> <p>You can always write these simply as polynomial maps of affine space, and then they are special cases of general theorems about polynomial maps of affine spaces. So presumably what you want to know is if you can use the matrix formulation to glean additional information. Have you looked at the $n=2$ case. Writing <code>$X=\left(\begin{smallmatrix} x&amp;y\\ z&amp;w\\ \end{smallmatrix}\right)$</code> and <code>$C=\left(\begin{smallmatrix} a&amp;b\\ c&amp;d\\ \end{smallmatrix}\right)$</code>, the map <code>$X\mapsto X^2+C$</code> is simply the map <code>$$F : \mathbb{A}^4\to\mathbb{A}^4,\qquad (x,y,z,w) \mapsto (x^2+yz+a, xy+wy+b, xz+wz+c, w^2+yz+d).$$</code> Homogenizing gives $\bar F : \mathbb{P}^4\to\mathbb{P}^4$, <code>$$\bar F(x:y:z:w:t) = (x^2+yz+at^2: xy+wy+bt^2: xz+wz+ct^2: w^2+yz+dt^2 : t^2)$$</code> The indeterminacy locus on the hyperplane at infinity is the rational curve parametrized by (if I've computed correctly) <code>$$\{ (uv, u^2,-v^2, -uv, 0) : u,v\in\mathbb{C} \}.$$</code></p> <p>What do you mean by "the" Mandelbrot set in this setting? I guess one could define it to be the points $C$ such that the orbits of the critical points are all bounded. Of course, the space of $C$ is not 4 dimensional, since you can always conjugate by an arbitrary invertible matrix. If you're working over $\mathbb C$, then you may assume that $C$ is in Jordan normal form, so the moduli space consists of two components, namely diagonal $C$ and $C$ consisting of a single non-semisimple Jordan block. You might consider each of these in turn. </p> http://mathoverflow.net/questions/131853/sequences-satisfying-gcdsx-sy-sgcdx-y/131865#131865 Answer by Joe Silverman for Sequences satisfying gcd(S(x), S(y)) = S(gcd(x,y)) Joe Silverman 2013-05-25T18:54:53Z 2013-05-25T18:54:53Z <p>A <em>strong divisibility sequence</em> is a sequence of positive integers $(a_n)_{n\ge1}$ with the property that $\gcd(a_n,a_m)=a_{\gcd(n,m)}$. See <a href="http://en.wikipedia.org/wiki/Divisibility_sequence" rel="nofollow">http://en.wikipedia.org/wiki/Divisibility_sequence</a>. (Strong) divisibility sequences tend to arise from algebraic groups, including sequences such as $a^n-b^n$, Fibonacci (and similar) sequences, elliptic divisibility sequences, etc. I'm not sure about your second condition. For elliptic divisibility sequences, there's a large literature on which terms are divisible by $p$, but it generally won't be the $(p-2)$'nd term. See <a href="http://en.wikipedia.org/wiki/Elliptic_divisibility_sequence" rel="nofollow">http://en.wikipedia.org/wiki/Elliptic_divisibility_sequence</a>, especially the section on EDS over finite fields. </p> http://mathoverflow.net/questions/131632/can-one-bound-the-quadratic-points-on-curves/131655#131655 Answer by Joe Silverman for Can one bound the Quadratic Points on Curves? Joe Silverman 2013-05-23T21:33:19Z 2013-05-23T21:33:19Z <p>I was going to point out that the specific result you cite for quadratic points is actually a theorem of Joe Harris and mine (Proc. Amer. Math. Soc. 112 (1991), 347-356), but I see that stankewicz has already given the reference, so I won't repeat the link. What Abramovich and Harris did later (and this is much harder) is generalize the result to points of degree $d$ for all $d\ge2$, although the obvious generalization turns out to be false, one can't merely assume that there are no maps of degree at most $d$ to an elliptic curve or to $\mathbb P^1$. Anyway, as far as I know, all finiteness results of this sort rely on a theorem of Faltings' (generalizing Vojta's proof of the Mordell conjecture) that is highly noneffective. However, for what it's worth, it is possible to give an effective upper bound for <code>$\#D_C$</code>. More generally, one can give effective constants $K_1$ and $K_2$ such that <code>$\#D_C$</code> has at most $K_1$ points whose height is greater than $K_2$. Unfortunately, we can't get our hands on those putative $K_1$ points of large height, although one suspects that there aren't any such points.</p> http://mathoverflow.net/questions/130868/ramification-in-division-field-of-abelian-varieties/130873#130873 Answer by Joe Silverman for Ramification in Division field of Abelian Varieties Joe Silverman 2013-05-16T20:33:43Z 2013-05-16T20:33:43Z <p>What if $m=pq$ with $\mathfrak p \mid p$ and $p\ne q$ and $k(A[p])=k$? Then the $p$-torsion doesn't cause ramification since its defined over $k$, and the $q$-torsion won't cause $\mathfrak p$ ramification (assuming $A$ has good reduction at the primes lying over $p$ and $q$).</p> <p>It gets more interesting if you assume that $m$ is a power of $p$.</p> http://mathoverflow.net/questions/129665/short-basis-for-the-unit-group-of-a-number-field/129678#129678 Answer by Joe Silverman for Short basis for the unit group of a number field Joe Silverman 2013-05-04T21:07:18Z 2013-05-04T21:07:18Z <p>The geometry of numbers should say that there is a basis for the units whose lengths are bounded by an explicit function of the regulator $R_K$. So this reduces to the question of upper bounds for $R_K$, and of course $R_K$ does not depend on the choice of basis. </p> <p>Since $R_K$ is logarithmic, one might hope for a bound of the form $(\log D_K)^t$, where $D_K$ is the absolute discriminant, but this is almost certainly false. For example, take a real quadratic field. It is conjectured that there are infinitely many with class number 1, so the fact that $\log (h_K R_K) \sim \frac12\log D_K$ says that $R_K$ is roughly on the order of $\sqrt{D_K}$. </p> <p>So in general, since $h_K\ge1$, one gets $R_K\le D_K^{\frac12+\epsilon}$ if you range over fields with $(\log D_K)/[K:\mathbb{Q}]\to\infty$. Presumably there are explicit and effective bounds if you're willing to accept a weaker estimate.</p> <p>Finally, in the other direction, one has lower bounds of the form (constants depend on the degree of the field) $$R_K \gg (\log D_K)^{r(K)-\rho(K)},$$ where $r(K)$ is the rank of the unit group of $K$, and $\rho(K)$ is the maximum of $r(k)$ as $k$ ranges over all proper subfields of $K$. For estimates of this sort, see</p> <ol> <li><p>M. Pohst, Regulatorabschätzungen für total reelle algebraische Zahlkörper, <em>J. Number Theory</em>, <strong>9</strong> (1977), pp. 459–492</p></li> <li><p>R. Remak, Über Grössenbesiehungen zwischen Diskriminante und Regulator eines algebraischen Zahlkörpers, <em>Compositio Math.</em>, <strong>10</strong> (1952), pp. 245–285</p></li> <li><p>J. Silverman, An inequality relating the regulator and the discriminant of a number field, <em>J. Number Theory</em>, <strong>19</strong> (1984), pp. 437–442</p></li> </ol> http://mathoverflow.net/questions/129509/elliptic-curve-with-a-degree-2-isogeny-to-itself/129515#129515 Answer by Joe Silverman for elliptic curve with a degree 2 isogeny to itself? Joe Silverman 2013-05-03T10:51:04Z 2013-05-03T10:51:04Z <p>Expanding on Francois's answer, $E$ has an endomorphism of degree 2 if and only if its endomorphism ring $R=\operatorname{End}(E)$, which is an order in an imaginary quadratic field, has an element of norm 2. There are exactly three such orders, namely $\mathbb{Z}[i]$, $\mathbb{Z}[\sqrt{-2}]$, and $\mathbb{Z}[(1+\sqrt{-7})/2]$. So up to isomorphism over $\overline{\mathbb{Q}}$, there are exactly three elliptic curves with endomorphisms of degree 2. Equations for these curves and their degree 2 endomorphism are given in <em>Advanced Topics in the Arithmetic of Elliptic Curves</em>, Proposition II.2.3.1.</p> <p>There are similarly only finitely many curves with a higher degree cyclic isogeny of fixed degree $d$. Using Velu's formulas, one could probably write them all down for small values of $d$.</p> http://mathoverflow.net/questions/128780/integer-dynamics-hitting-infinitely-many-primes/128794#128794 Answer by Joe Silverman for Integer dynamics hitting infinitely many primes Joe Silverman 2013-04-26T03:00:05Z 2013-04-26T03:00:05Z <p>As Mark says, nothing is known about infinitely many primes in dynamical sequences, other than ones that contain an arithmetic progression. An easier, but still useful, question, is that of primitive prime divisors. A prime $p$ is a <em>primitive prime divisor</em> of $f^n(x)$ if $p\mid f^n(x)$ and $p\nmid f^m(x)$ for all $m\lt n$. There's a vast literature on primitive prime divisors in various sorts of sequences. In general, if $\mathcal{A}=(a_n)$ is a sequence of integers, or more generally, rational numbers, the <em>Zsigmondy set</em> of $\mathcal{A}$ is <code>$$\mathcal Z(\mathcal A) = \{ n : \hbox{the numerator of a_n has no primitive prime divisors} \}.$$</code> Patrick Ingram and I [1] showed that under suitable hypotheses on $f\in\mathbb{Q}(T)$ and $x,y\in\mathbb{Q}$, the Zsigmondy set $\mathcal{Z}(f^n(x)-y)$ is finite. In addition to some obvious conditions needed to avoid trivial counterexamples, we needed to assume that $y$ is preperiodic for $f$. Recently, Gratton, Nguyen, and Tucker [2] removed this restriction on $y$, conditional on the $abc$ conjecture.</p> <p>[1] Ingram, Patrick; Silverman, Joseph H.; Primitive divisors in arithmetic dynamics. <em>Math. Proc. Cambridge Philos. Soc.</em> <strong>146</strong> (2009), no. 2, 289–302 (MR2475968)</p> <p>[2] Chad Gratton, Khoa Nguyen, Thomas J. Tucker, ABC implies primitive prime divisors in arithmetic dynamic, preprint, <a href="http://arxiv.org/abs/1208.2989" rel="nofollow">http://arxiv.org/abs/1208.2989</a></p> http://mathoverflow.net/questions/128176/will-quantum-computing-kill-cryptography/128179#128179 Answer by Joe Silverman for Will quantum computing kill cryptography ? Joe Silverman 2013-04-20T17:00:58Z 2013-04-20T17:00:58Z <p>This question seems a bit vague, but one answer is that there are cryptosystems such as NTRU that are based on (special cases) of the closest vector problem (CVP). At present, quantum computers would not significantly speed up the solution of the CVP. If I understand correctly, they would require doubling the length of the keys. </p> <p>Disclaimer: Jeff Hoffstein, Jill Pipher, and I are the ones who devised NTRU. But there are other lattice-based systems out there (though generally not as efficient). In any case, I think a good answer to your question is that you should look at lattice-based cryptography for examples. </p> http://mathoverflow.net/questions/125664/division-by-3-on-elliptic-curve/125668#125668 Answer by Joe Silverman for Division by 3 on elliptic curve Joe Silverman 2013-03-26T21:19:17Z 2013-03-26T21:19:17Z <p>The fact that you quote comes from the Kummer sequence. Let $G_K=\text{Gal}(\bar K/K)$. Then one gets <code>$$E(K)/2E(K) \hookrightarrow H^1(G_K,E[2]) = \text{Hom}(G_K,\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}) \cong K^*/(K^*)^2 \times K^*/(K^*)^2.$$</code> This works because you're taking a curve for which all of the $2$-torsion is rational. You can do something similar if all of the $m$ torsion is rational, yielding <code>$$E(K)/mE(K) \hookrightarrow H^1(G_K,E[m]) = \text{Hom}(G_K,\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/m\mathbb{Z}) \cong K^*/(K^*)^m \times K^*/(K^*)^m.$$</code> (Note that the assumption that the $m$ torsion is rational implies that $K$ contains the $m$'th roots of unity, which is why the last isomorphism is okay.) From this one can in principle derive two functions $F$ and $G$ on $E$ with the property that $P$ is $m$ times a point if and only if $F(P)$ and $G(P)$ are $m$'th powers. But I don't know a reference offhand.</p> <p>This may be in Husemoller's book, or you can find it in Chapter X of my <em>Arithmetic of Elliptic Curves</em>.</p> http://mathoverflow.net/questions/123655/the-elliptic-lehmer-problem-for-several-independent-algebraic-points/123688#123688 Answer by Joe Silverman for The elliptic Lehmer problem for several independent algebraic points Joe Silverman 2013-03-06T00:39:00Z 2013-03-07T14:36:37Z <p>ADDENDUM: I looked at Masser's 1989 article, and a quick back-of-the-envelope calculation seems to give the result, at least for two points. Thus if $P$ and $Q$ are independent points generating a field of degree $d$, then $$\hat h(P)\hat h (Q) \ge \frac{C(E)}{d^3(\log d)^2}.$$ The idea is to apply the main theorem in Masser's paper to the set of points <code>$$\{ mP + nQ : 0\le m\lt M~\text{and}~0\le n\lt N \}.$$</code> Then choose the optimal values for $M$ and $N$, much as one does in proving the Cauchy-Schwarz inequality. Presumably one can do something similar with more points.</p> <hr> <p>@Jamie: This is not Lang's height conjecture, which fixes a field (or degree of the field) and lets the elliptic curve vary. This is, as to OP indicates, a generalization of the elliptic version of Lehmer's conjecture. </p> <p>@Vessilin: I remember corresponding with someone (maybe Cam Stewart) at one point about whether one could get a better bound for $\max \hat h(P_i)$, but we never got an interesting result. And I don't know anything about the problem of minimizing a product. Looks like an interesting problem. There are, as far as I know, three basic approaches that have been used for the $r=1$ case.</p> <p>(1) For the CM case, Michel Laurent's estimate that mirrors Dobrowolski's.</p> <p>(2) Use of Diophantine approximation / transcendence theory methods (i.e., auxiliary polynomials that become so small, they must vanish). The best estimate (up to improving the constant) is still Masser's and gives something like $\hat h(P) \ge C(E)/d^3(\log d)^2$. </p> <p>(3) Use of Fourier averaging (which is the same idea that Hindry and I used for Lang's height conjecture). Hindry and I could only make this work if the $j$-invariant of $E$ is <em>non-integral</em>, but with this added hypothesis, we knocked one off of Masser's exponent and got $\hat h(P) \ge C(E)/d^2(\log d)^2$. </p> <p>The reason I mention these is that it seems reasonable (to me) that each method has at least the possibility of being used to obtain a result for $\hat h(P_1)\hat h(P_2)\cdots\hat h(P_r)$.</p> <p>There is also the following slightly different result: If you only look at points defined over the maximal abelian extension $K^{ab}$ of $K$, then there is an absolute upper bound $\hat h(P)\ge C(E)$. This was proven by combining two article, one by Matt Baker and one of mine, and then we have a joint paper that generalized it to abelian varieties. It is based on the ideas of Amoroso and Dvornicich, who proved the analogous result for the multiplicative group. The method can be roughly described as a careful use of ramification. Of course, in this case a product would have the obvious lower bound $C(E)^r$, so not very interesting. But I mention it since there are so few known methods that are potentially useful.</p> http://mathoverflow.net/questions/123352/effective-mordell/123374#123374 Answer by Joe Silverman for Effective Mordell Joe Silverman 2013-03-01T20:58:51Z 2013-03-01T20:58:51Z <p>If one had an <em>effective</em> version of the abc conjecture, then Elkies showed how to use it to obtain an effective version of the Mordell conjecture (using Belyi maps).</p> <p>In your formulation, the theorem would be effective if there was an algorithm to compute the constant const_f for any given f. </p> <p>For example, assume that $f$ has coefficients in $\mathbb Z$, and let $H(f)$ be the maximum of the absolute values of its coefficients. Then the following would be an effective version of the Mordell conjecture: $$\max(|n|,|m|,|n'|,|m'|) \le 10^{10^{10^{H(f)+\deg(f)+1000}}}.$$ NOTE: I'm not saying that this statement is known; it's not. (Although I'd be surprised if it isn't true.) But it illustrates what is meant by an "effective bound".</p> http://mathoverflow.net/questions/120612/trichotomies-in-mathematics/120665#120665 Answer by Joe Silverman for Trichotomies in mathematics Joe Silverman 2013-02-03T13:21:22Z 2013-02-03T13:21:22Z <p>The reduction (special fiber) $E_{\mathfrak p}$ of (the Neron model of) an elliptic curve $E$ modulo a prime ${\mathfrak p}$ is one of:</p> <ol> <li>good reduction = stable reduction = $E_{\mathfrak p}$ is non-singular</li> <li>multiplicative reduction = semi-stable reduction = $E_{\mathfrak p}$ is a product of the multiplicative group times a finite group</li> <li>additive reduction = unstable reduction = $E_{\mathfrak p}$ is a product of the additive group times a finite group</li> </ol> <p>Of course, this trichotomy is a reflection of the fact that there are only three sorts of connected one-dimensional Lie groups, namely the additive group, the multiplicative group, and the compact case (elliptic curves).</p> http://mathoverflow.net/questions/120317/the-height-of-an-orbit-under-rational-self-maps/120379#120379 Answer by Joe Silverman for The height of an orbit under rational self-maps Joe Silverman 2013-01-31T04:01:54Z 2013-01-31T04:01:54Z <p>I thank Mahdi for the pointer to the paper. It was my first paper on this subject. It considers especially the case of monomial maps. It contains some references to a small number of papers by other people who have studied the growth rate of $h(\phi^n(x_0))$.</p> <p>Dynamical Degrees, Arithmetic Degrees, and Canonical Heights for Dominant Rational Self-Maps of Projective Space, <a href="http://arxiv.org/abs/1111.5664" rel="nofollow">http://arxiv.org/abs/1111.5664</a></p> <p>In the general setting $\phi:X\to X$, let $x_0$ be a point whose entire forward orbit is well-defined. A somewhat coarse, but still quite interesting, measure of the growth rate of $h(\phi^n(x_0))$ is the <em>arithmetic degree</em>, which by definition is the limit (if the limit exists) $$\alpha_\phi(x_0) = \lim_{n\to\infty} h(\phi^n(x_0))^{1/n}.$$ Shu Kawaguchi and I have studied the arithmetic degree, and its relation to the geometrically defined dynamical degree $\delta_\phi$ of $\phi$, in several papers. For example, we proved in general that $\alpha_\phi(x_0)\le\delta_\phi$ (this is easy for projective space, but gets harder if the Neron-Severi group has rank larger than 1), we proved that for morphisms, the limit defining $\alpha_\phi(x_0)$ always exists and is an algebraic integer, and we proved that if $X=E^N$ is a power of a non-CM elliptic curve and $\phi$ is an isogeny and the orbit of $x_0$ is Zariski dense, then there is equality $\alpha_\phi(x_0)=\delta_\phi$. (We conjecture these properties are true in general.)</p> <p>Here are the links. All are joint with Shu Kawaguchi.</p> <p>On the dynamical and arithmetic degrees of rational self-maps of algebraic varieties, <a href="http://arxiv.org/abs/1208.0815" rel="nofollow">http://arxiv.org/abs/1208.0815</a></p> <p>Examples of dynamical degree equals arithmetic degree, <a href="http://arxiv.org/abs/1212.3015" rel="nofollow">http://arxiv.org/abs/1212.3015</a></p> <p>Dynamical Canonical Heights for Jordan Blocks and Arithmetic Degrees of Orbits, <a href="http://arxiv.org/abs/1301.4964" rel="nofollow">http://arxiv.org/abs/1301.4964</a></p> <p>Regarding the statement in your edit, are you using anything more than the fact that the orbit is contained in $X(K)$ for some number field $K$, and then apply the trivial upper bound for the number of $K$-rational points in projective space? I think that statements of this sort can be interesting, but probably the $o(\log n)$ needs to be replaced by a function that grows significantly faster to get a really interesting statement. </p> http://mathoverflow.net/questions/111980/reference-request-for-the-theory-of-heights-over-function-fields/119512#119512 Answer by Joe Silverman for Reference request for the theory of heights over function fields Joe Silverman 2013-01-21T20:40:32Z 2013-01-21T20:40:32Z <p>This is discussed (using somewhat older language) in Lang's <em>Fundamentals of Diphantine Geometry</em>. See in particular Chapter 3, Section 3, which is called "Heights in Function Fields". There is a theorem of Neron (Theorem 3.6 on page 66) which says that bounded height implies that the associated map has bounded degree. Then one should be able to use the theory Hilbert schemes (or as Lang uses in Chapter 6, the theory of Chow coordinates) to complete the argument. See also the discussion in Chapter 6, Section 5, which is entitled "Points of Bounded Height", where he uses Neron's result to analyze points of bounded height on abelian varieties; it seems that at least parts of the argument should apply generally.</p> http://mathoverflow.net/questions/117241/arithmetic-dynamics-and-dynamics-on-moduli-spaces/117282#117282 Answer by Joe Silverman for Arithmetic dynamics and dynamics on moduli spaces Joe Silverman 2012-12-26T21:11:51Z 2012-12-26T21:11:51Z <p>In addition to the graduate textbook that Felipe already mentioned, </p> <p><em>The Arithmetic of Dynamical Systems</em>, Springer-Verlag, GTM <strong>241</strong>, 2007,</p> <p>there's also the following monograph that discusses dynamical-related moduli spaces from an algebraic and arithmetic viewpoint:</p> <p><em>Moduli Spaces and Arithmetic Dynamics</em>, CRM Monograph Series <strong>30</strong>, AMS, 2012.</p> <p>As for the list of references <a href="http://www.math.brown.edu/~jhs/ADSBIB.pdf" rel="nofollow">http://www.math.brown.edu/~jhs/ADSBIB.pdf</a> that Benjamin Dickman mentioned, I update it once or twice a year, so it's reasonably up-to-date, but I make no claim to its being complete. You might also try searching MathSciNet and the ArXiv for articles whose classification number is 37Pxx, which will catch most recent articles. In particular, there's</p> <p>37P45 = Dynamical systems and ergodic theory / Arithmetic and non-Archimedean dynamical systems / Families and moduli spaces.</p> http://mathoverflow.net/questions/116318/elementary-proof-of-mordells-theorem/116620#116620 Answer by Joe Silverman for Elementary proof of Mordell's theorem Joe Silverman 2012-12-17T16:49:23Z 2012-12-17T16:49:23Z <p>Not sure if this answers your question, but here's a thought. </p> <p>The known proofs that $E(K)/mE(K)$ is finite are non-effective because they embed $E(K)/mE(K)$ into a larger group that is shown to be finite. Let's call that larger group $S^{(m)}$. The proof that $S^{(m)}$ is finite is effective, and indeed it yields a very nice upper bound for the order of $S^{(m)}$. However, as far as I know, this bound always involves (at least) the $m$-part of the class number of $K(P)$, where $P$ is an $m$-torsion point of $E$. So the proof requires algebraic number theory in the sense that one needs to know that the class group (or at least the $m$ part) of some extension field of $K$ is finite, unless there is a rational $m$ torsion point. So proofs along these lines would seem to require algebraic number theory and ideal class groups, either explicitly or disguised in some way.</p> http://mathoverflow.net/questions/115112/how-does-modern-number-theory-contribute-to-further-understanding-of-mathbbn/115119#115119 Answer by Joe Silverman for How does "modern" number theory contribute to further understanding of $\mathbb{N}$? Joe Silverman 2012-12-01T22:00:19Z 2012-12-01T22:00:19Z <p>Where to begin? As a tiny example, suppose that you are interested in the solutions in rational numbers to an equation $Y^2=X^3+AX+B$, i.e., the rational points on an elliptic curve $E$. One naturally looks at the algebraic solutions $E(\overline{\mathbb{Q}})$, since one can do geometry over the algebraically closed field $\overline{\mathbb{Q}}$, and then picks out the rational points $E(\mathbb{Q})$ by studying the action of the Galois group $G(\overline{\mathbb{Q}}/\mathbb{Q})$ and picking out the points that are invariant for the group action. Hopefully you're also interested in rational numbers, since they're just ratios of integers, but if you insist on problems with integers, then one can look at the integer solutions $E(\mathbb{Z})$, which forms a finite set (Siegel's theorem, made effective by Baker). But even there, one way to study $E(\mathbb{Z})$ is by analyzing it as a subset of $E(\mathbb{Q})$, so you're back to rational solutions.</p> <p>So that's a bit long-winded, but it illustrates a general principle. If a problem involving a set is hard, for example a problem involving integers, it may be easier to solve a problem involving a larger set and then pick out the subset that you're really interested in.</p> <p>Having said all of this, it is also true that there are mathematicians who find studying "new" objects to be intrinsically interesting, regardless of the original applications that they were devised for. As an example, there are lots of people who study automorphic forms for their own sake. If there are applications to classical sorts of problems involving integers, that's great, but it's not their motivation or primary interest. Luckily, there's room for everyone in the big tent that makes up the mathematical research community.</p> http://mathoverflow.net/questions/109298/example-of-a-diophantine-application-of-an-open-image-theorem/109348#109348 Answer by Joe Silverman for Example of a diophantine application of an open image theorem Joe Silverman 2012-10-11T02:02:25Z 2012-10-11T02:02:25Z <p>Here's an application to independence of Heegner points. (But if you search on MathSciNet for papers that reference Serre's two results, I expect you'll find a very large number of applications.)</p> <p>Let $E/\mathbb{Q}$ be an elliptic curve with no CM, and let $\Phi:X_0(N)\to E$ be a modular parametrization. (Wiles et.al. show that $\Phi$ exists for all such $E$.) The modular curve $X_0(N)$ has special points called <em>Heegner points</em> associated to pairs $(C,\Gamma)$, where $C$ is a CM elliptic curve and $\Gamma\subset C$ is a cyclic subgroup of order $N$. More precisely, we can associate to each imaginary quadratic field $K$ (satisfying some conditions) a Heegner point $x_K\in X_0(\overline{\mathbb{Q}})$ associated to the maximal order in $K$.</p> <p><strong>Theorem</strong> [1] Let $K_1,\ldots,K_r$ be distinct imaginary quadratic fields such that the odd parts of their class numbers are sufficiently large. Then the points $\Phi(x_{K_1}),\ldots,\Phi(x_{K_r})$ are linearly independent in the group $E(\overline{\mathbb{Q}})$.</p> <p>The proof uses Serre's image of Galois theorem in a crucial way. Not simply that the image of Galois is open in each $\hbox{Aut}(T_\ell(E))$, but also that it is surjective for almost all $\ell$.</p> <p>[1] M. Rosen, JH Silverman, On the independence of Heegner points associated to distinct quadratic imaginary fields, <em>Journal of Number Theory</em> <strong>127</strong> (2007), 10-36.</p> http://mathoverflow.net/questions/109250/another-question-related-to-the-isogeny-theorem-for-elliptic-curves/109256#109256 Answer by Joe Silverman for Another question related to the isogeny theorem for elliptic curves Joe Silverman 2012-10-09T23:09:10Z 2012-10-09T23:09:10Z <p>I think that maybe what is meant is that there is no <em>functorial</em> way to define the bijection in the category of algebraic geometry. So suppose that we let $\hbox{Ell}(R)$ denote the set of elliptic curves with $\hbox{End}(E)\cong R$, where for simplicity $R$ is the maximal order of an imaginary quadratic field. (The isomorphism with $R$ is over $\overline{\mathbb{Q}}$.) The ideal class group $H_R$ is, as its name proclaims, a group. So it has a preferred element, namely the identity element. But $\hbox{Ell}(R)$ does not have a preferred element in any natural sense. The right way to think of this is that there is a natural action of $H_R$ on $\hbox{Ell}(R)$, and this action is simply transitive. In particular, $H_R$ and $\hbox{Ell}(R)$ have the same number of elements. But if you want to identify them $\hbox{Ell}(R)\leftrightarrow H_R$, you need to choose an element of $\hbox{Ell}(R)$ to be distinguished.</p> <p>On the other hand, if you fix an embedding $R\subset\mathbb{C}$, then every ideal $\mathfrak{a}$ in $R$ is a lattice in $\mathbb{C}$, so you can identify the ideal class $\overline{\mathfrak{a}}$ with the complex torus $\mathbb{C}/\mathfrak{a}$. This is analytically isomorphic to an elliptic curve $E_{\mathfrak{a}}$ in $\hbox{Ell}(R)$.</p> http://mathoverflow.net/questions/108726/getting-a-bound-on-the-coefficients-of-the-factor-polynomial/108761#108761 Answer by Joe Silverman for Getting a bound on the coefficients of the factor polynomial Joe Silverman 2012-10-03T22:47:12Z 2012-10-03T22:47:12Z <p>Gelfand's inequality is probably what you want here; see for example my book with Hindry, <em>Diophantine Geometry: An Introduction</em>, Proposition B.7.3. I'll state it for polynomials in $\mathbb{Z}[X_1,\ldots,X_m]$, although there's a version that's true over $\overline{\mathbb{Q}}$. The statement uses the projective height, so for a polynomial $f$ with coefficients $a_i\in\mathbb{Z}$, we let $$H(f) = \frac{\max|a_i|}{\gcd(a_i)}.$$ Then</p> <p><strong>Proposition B.7.3 (Gelfand's inequality)</strong> Let $f_1,\ldots,f_r\in \mathbb{Z}[X_1,\ldots,X_m]$, and for $1\le i\le m$, let $d_i$ denote the $X_i$ degree of $f_1f_2\cdots f_r$. Then $$H(f_1)H(f_2)\cdots H(f_r) \le e^{d_1+\cdots+d_m}H(f_1f_2\cdots f_r).$$</p> <p>For the OP's question, we have $f$ is divisible by $g$, say $f=gg'$, so $$H(g) \le H(g)H(g') \le e^{\deg f}H(gg') = e^{\deg f}H(f).$$</p> http://mathoverflow.net/questions/107708/origin-of-square-and-multiply-algorithm Origin of square-and-multiply algorithm Joe Silverman 2012-09-20T19:19:57Z 2012-09-22T10:01:29Z <p>I'm teaching an introductory course in cryptography and explained the square-and-multiply algorithm to the class. </p> <p><a href="http://en.wikipedia.org/wiki/Square-and-multiply_algorithm" rel="nofollow">http://en.wikipedia.org/wiki/Square-and-multiply_algorithm</a></p> <p>Someone asked who discovered the algorithm, which I didn't know, so after a short web search that gave no answers, I thought I'd ask on MO. In particular, the above wikipedia article is not helpful, and I didn't see any MO questions that address the issue. This seems like something that Gauss and Euler, or even Fermat, might have known, and ditto for Indian and Chinese mathematicians centuries earlier, but I'm just speculating. Specific references would be appreciated. (Sorry if this isn't really a research level question, although maybe it qualifies as historical research.)</p> http://mathoverflow.net/questions/107287/finiteness-of-elliptic-curves-of-a-given-conductor/107289#107289 Answer by Joe Silverman for Finiteness of elliptic curves of a given conductor Joe Silverman 2012-09-15T22:03:20Z 2012-09-15T22:03:20Z <p>As Noam says, this was well-known long before Wiles. The proof that he sketches can be used to prove Shafarevich's theorem for elliptic curves, i.e., given a finite set of primes $S$, there are only finitely many elliptic curves over $\mathbb{Q}$ with good reduction outside of $S$. So if you have a particular $N$ in mind, you can take $S$ to be the set of primes dividing $N$.</p> <p>And there were even some fairly explicit upper bounds for the number of elliptic curves of conductor $N$. See for example my paper with Brumer, "The number of elliptic curves over $\mathbb{Q}$ with conductor $N$," <em>Manuscripta Math.</em> <strong>91</strong> (1996), 95-102. Turning things around, we can then use Wiles' theorem and the previously proven upper bound for the number of elliptic curves of conductor $N$ to immediately deduce a non-trivial upper bound for the number of elliptic factors of $J_0(N)$.</p> http://mathoverflow.net/questions/2876/behaviour-of-zeta-function-under-finite-morphism/106008#106008 Answer by Joe Silverman for Behaviour of Zeta-function under Finite Morphism Joe Silverman 2012-08-31T00:03:21Z 2012-08-31T00:03:21Z <p>Let $E\to C$ be an elliptic surface (say defined over $\mathbb{Q}$. Then Mike Rosen and I proved a formula for the rank of the group of sections $C\to E$ defined over $\mathbb{Q}$ in terms of the average number of points modulo $p$ on the fibers. (The result is conditional on Tate's conjecture for the surface $E$.) Anyway, the proof is via a comparison of the zeta function of $E$ as a surface compared to the zeta functions of the fibers and the zeta function of the base curve $C$. So not a direct relation that you seem to be asking for, but it has the flavor of your question. The paper is</p> <p>M. Rosen, J.H. Silverman, On the rank of an elliptic surface, Invent. Math. 133 (1998), 43-67.</p> <p>There have since been some extensions to 1-dimensional families of abelian varieties and/or to higher dimensional bases, which you can find by forward referencing our article on MathSciNet.</p> http://mathoverflow.net/questions/104066/proving-finite-generation-by-tensoring-with-mathbbr/104110#104110 Answer by Joe Silverman for Proving finite generation by tensoring with $\mathbb{R}$ Joe Silverman 2012-08-06T13:52:26Z 2012-08-06T13:52:26Z <p>How do you prove that the ring of integers in a number field is finitely generated? One usually embeds them in $\mathbb{R}^{r_1}\times\mathbb{C}^{r_2}$. How do you prove that the units in a number field are finitely generated? One generally embeds them in a hyperplane in $\mathbb{R}^{r_1+r_2-1}$. Then one shows that the group sits as a discrete subgroup, hence is finitely generated. Further, by looking at the co-volume of the resulting lattice, one obtains important arithmetic invariants, namely the discriminant and the regulator. Anyway, one can call almost any technique a trick, but the idea of embedding a group into a real or complex vector space and using volume estimates to prove discreteness is quite a well-established technique. And if one simply has the group and a positive definnite quadratic form, as in the case of End$(E)$ or the canonical height on $E(\mathbb{Q})$, then it is very natural to tensor with $\mathbb{R}$ and extend the quadratic form to put a Euclidean structure on the resulting vector space. </p> <p>Of course, as Will Sawin says, one can alternatively view the embedding as being into the dual. But for example, using the height pairing on $E(\mathbb{Q})/tors$, the values of $\hat h$ are real, so even though $E(\mathbb{Q})/tors$ is a torsion-free $\mathbb{Z}$-module, the dual space that one uses has to be Hom$(E(\mathbb{Q})/tors,\mathbb{R})$.</p> <p>I don't know that this really answers the question, but I hope it gives some indication of why it's not so unusual to think of embedding $M$ into a real vector space as the first step in proving that it is finitely generated. </p> <p>Finally, I should mention that I first saw the argument for proving that End$(E)$ is finitely generated in Mumford's <em>Abelian Varieties</em>, but I don't know the origins of the idea.</p> http://mathoverflow.net/questions/76413/torsion-subgroups-in-families-of-twists-of-elliptic-curves/76464#76464 Answer by Joe Silverman for Torsion subgroups in families of twists of elliptic curves Joe Silverman 2011-09-27T03:15:29Z 2012-08-06T13:28:24Z <p><strong>Theorem</strong> (originally due to Setzer?): Fix $E/\mathbb{Q}$ with $j(E)$ not 0 or 1728. Then for all but finitely many inequivalent twists $E_d$, the torsion subgroup $E_d(\mathbb{Q})_{tors}$ is isomorphic to <code>$E[2](\mathbb{Q})$</code>, so in particular $E_d(\mathbb{Q})_{tors}$ has order 1, 2, or 4. (Probably he also proved it for number fields.)</p> <p>There's a paper of mine$^1$ with a much more general theorem using the theory of heights. I don't recall Setzer's proof except that it doesn't use heights.</p> <p><strong>Theorem</strong>: Let $K$ be a number field and let $A/K$ be an abelian variety with $\mu_n\subset {\rm Aut}(A)$. (This means we can twist $A$ by $n$'th roots of $d$.) Then every point $P\in A_d(K)$ satisfies one of the following two conditions:</p> <ul> <li>$P$ is fixed by a non-trivial $\zeta\in\mu_n$.</li> <li>$\hat h(P) \ge C_1(A)h^{(n)}(d) - C_2(A)$.</li> </ul> <p>Here $\hat h$ is the canonical height relative to an ample symmetric divisor, and $h^{(n)}(d)$ is a sort of "$n$'th power free height," say equal to the minimum of $h(du^n)$ for $u\in K^*$. The constants depend on $A$, but are independent of $d$.</p> <p>It follows from the theorem that after discarding finitely many <code>$d \in K^*/{K^*}^n$</code>, a point in $A_d(K)$ is either $1-\zeta$ torsion (hence $nP=O$), or its canonical height is positive, and hence it is nontorsion.</p> <p>Of course, to describe more precisely what happens for the finitely many exceptional $d$ can be a delicate matter, as some of the other answers have indicated. I think it's interesting to see how one can approach the problem via heights or via representation theory.</p> <p>$^1$ J.H. Silverman, Lower bounds for height functions, <em>Duke Math. J.</em> <strong>51</strong> (1984), 395-403.</p> <p>EDIT: Fixed statement of first theorem. I'd originally written that "for all but finitely many inequivalent twists $E_d$, the torsion subgroup $E_d(\mathbb{Q})_{tors}$ has at most two elements." This is clearly false, since if $E$ has the form $E:y^2=(x-a)(x-b)(x-c)$ with $a,b,c\in\mathbb{Q}$, then <code>$E_{d}[2](\mathbb{Q})$</code> has order 4 for every twist.</p> http://mathoverflow.net/questions/103308/overview-of-arakelov-intersection-theory-and-the-arakelov-chow-ring Overview of Arakelov intersection theory and the Arakelov Chow ring Joe Silverman 2012-07-27T13:40:19Z 2012-07-27T14:31:12Z <p>I'm looking for a reference that gives an overview of the most important properties of Arakelov intersection theory (on arithmetic varieties of arbitrary dimension) and that describes basic properties of the Arakelov Chow ring. There's a similar MO question asking about <a href="http://mathoverflow.net/questions/52665/survey-article-on-intersection-theory" rel="nofollow">survey articles on (classical) intersection theory</a>, so I guess that I'm asking the same question, but for Arakelov intersection theory.</p> http://mathoverflow.net/questions/102765/existence-of-an-elliptic-curve-and-a-distortion-map/102792#102792 Answer by Joe Silverman for Existence of an elliptic curve and a distortion map Joe Silverman 2012-07-21T03:08:13Z 2012-07-21T03:08:13Z <p>I assume $n$ is $m$. Also, of course, the map $\phi$ is supposed to be an isogeny.</p> <p>It's late, so this may not be quite right, but here's a thought. Take $E$ to have endomorphism ring $\mathbb{Z}[\sqrt{-D}]$. We at least want to choose $D$ so that the endomorphism $\phi=\sqrt{-D}$ does not behave like multiplication by an integer on $E[m]$. So we want $\sqrt{-D}\not\equiv a\pmod{m}$ for <em>all</em> integers $a$. It suffices to require that the norm of $a+\sqrt{-D}$ not be divisible by $m$. So we're looking for an integer $D$ such that $a^2+D\not\equiv0\pmod{m}$ for all integers $a$. It suffices to choose $D$ so that $-D$ is not a square modulo some prime dividing $m$.</p> <p>This will at least get you a map so that if $T$ has order $m$, then the group generated by $T$ and $\phi(T)$ is strictly bigger than $m$. I'm pretty sure that with a little more work (choosing $D$ more carefully), you can get $T$ and $\phi(T)$ to generate $E[m]$, but I'll stop here.</p> http://mathoverflow.net/questions/19930/writing-papers-in-pre-latex-era/102097#102097 Answer by Joe Silverman for Writing papers in pre-LaTeX era? Joe Silverman 2012-07-13T03:13:20Z 2012-07-13T03:13:20Z <p>I'm feeling old (like Felipe). In 1981 I typed my thesis on an IBM Selectric, typing the text on each page, then reinserting it and using the symbol typeball to fill in the symbols. And as an added handicap, I typed much of it with my daughter on my lap (from age 4 months to 6 months). Luckily, my arms are long, so I could keep her far enough from the keyboard that at worst she occasionally managed to press the space bar.</p> <p>A few years later, my first book was typed by a wonderful technical typist at MIT, after which it was typeset in Hong Kong. I ended up proof reading it (at least) 8 times, twice for the handwritten version, twice for the typed version, twice for the galley proofs, twice for the page proofs. So with 8 proofreadings, after it appeared, there were only about 400 typos (in a 400 page book), some of which I went back and checked, and sure enough, they were there in the original handwritten version. Sigh...</p> <p>I know this thread isn't supposed to be about TeX, but I have to mention what a thrill it was to be able to type my <em>Advanced Topics...</em> book on a MacPlus and have the processed TeX (plain, not LaTeX) appear on my screen in only 10 seconds per page. Quite a change over the years, since now all 500 pages of that book take less than 10 seconds on my laptop.</p> http://mathoverflow.net/questions/133834/mersenne-prime-sequences/133967#133967 Comment by Joe Silverman Joe Silverman 2013-06-17T18:35:48Z 2013-06-17T18:35:48Z I think that alias is overstating what's been proven. There are <i>certain</i> elliptic divisibility sequences and <i>certain</i> polynomial orbits for which one can prove there are only finitely many primes. Some of these, in particular the ones that Graham Everest and his colleagues/students studied, can be somewhat subtle to prove; but roughly speaking, they all come from cases where there is a &quot;map&quot; from some other sequence that provides a nontrivial divisor. Of course, there are also trivial examples. But in general, it is only <i>conjectured</i> that EDS and poly orbits have finitely many primes. http://mathoverflow.net/questions/133852/are-there-varieties-such-that-every-model-of-theirs-has-a-rational-point Comment by Joe Silverman Joe Silverman 2013-06-16T01:32:33Z 2013-06-16T01:32:33Z Probably I'm missing something here, but isn't it true that if $X/\mathbb{Q}$ has no $\overline{\mathbb{Q}}$ automorphisms, then it has a unique model over $K$? So it would suffice to take such a variety with $X(\mathbb{Q})\ne\emptyset$, since then $X_K=X_{\mathbb{Q}}\times\mbox{Spec}(K)$ will have the point inherited from $X(\mathbb{Q})$. OTOH, if you assume that there are non-trivial automorphisms, then it seems like an interesting question, since there will be number fields over which there are lots of non-trivial twists. http://mathoverflow.net/questions/132011/can-repunits-be-perfect-cubes/132170#132170 Comment by Joe Silverman Joe Silverman 2013-05-29T03:16:48Z 2013-05-29T03:16:48Z Hi Noam. Your &quot;strange step&quot; reminds me of V.A. Lebsgue's solution of y2=x3+7, although it's not as elaborate. First note x must be odd, else 7 would be a square mod 8. Then add 1 to get $$y^2+1=x^3+8=(x+2)(x^2-2x+4)=(x+2)((x-1)^2+3).$$ The last factor is 3 mod 4, so is divisible by a prime $q\equiv3\pmod4$. Then $y^2+1\equiv0\pmod{q}$, contradicting the fact that $-1$ is not a square mod $q$. So similar to your solution, he uses only the computation of $(-1|q)$. I don't know the exact reference, but I think that it was late 19th century, so it's not a new idea. http://mathoverflow.net/questions/131353/help-with-this-system-of-diophantine-equations/131361#131361 Comment by Joe Silverman Joe Silverman 2013-05-21T20:57:57Z 2013-05-21T20:57:57Z It's always intrigued me that for a given $m$, solving $a^3-b^3=m$ is easy (at least if we can factor $m$), but solving $a^3-2b^3=m$ is very difficult. An &quot;intrinsic&quot; explanation is that $\mathbb Z$ has only two units, while $\mathbb Z[2^{1/3}]$ has infinitely many units. But still, its amazing that there was no general effective solution method for $a^3-2b^3=m$ until Baker's theorem. http://mathoverflow.net/questions/130784/problem-factoring-a-cubic-polynomial Comment by Joe Silverman Joe Silverman 2013-05-16T02:41:53Z 2013-05-16T02:41:53Z Please read the FAQ. This site is for research level questions. Try MathStackExchange instead. http://mathoverflow.net/questions/129665/short-basis-for-the-unit-group-of-a-number-field Comment by Joe Silverman Joe Silverman 2013-05-04T21:09:22Z 2013-05-04T21:09:22Z @Asaf: Aren't real quadratic fields the 1-dimensional case? So your comments are probably more relevant to, say, a totally real cubic field. http://mathoverflow.net/questions/76413/torsion-subgroups-in-families-of-twists-of-elliptic-curves/129401#129401 Comment by Joe Silverman Joe Silverman 2013-05-02T14:15:41Z 2013-05-02T14:15:41Z More generally, it's easy to define a map $$E_d(\mathbb{Q}) \oplus E(\mathbb{Q}) \to E(\mathbb{Q}(\sqrt{d}))$$ whose kernel and co-kernel are $2$-groups. The result on $n$-torsion for odd $n$ is then immediate, and in addition one obtains the well-known and useful formula $$\operatorname{rank} E_d(\mathbb{Q}) + \operatorname{rank} E(\mathbb{Q}) \ = \operatorname{rank} E(\mathbb{Q}(\sqrt{d}))$$ http://mathoverflow.net/questions/128780/integer-dynamics-hitting-infinitely-many-primes Comment by Joe Silverman Joe Silverman 2013-04-26T02:40:33Z 2013-04-26T02:40:33Z @Anthony - Actually, in arithmetic dynamics your sequence 7,7,7,... is considered an arithmetic progression, with initial term 7 and common difference 0. This is a convenient convention, for example, for stating the dynamical Mordell-Lang conjecture, which says that <code>$\{n : f^n(x)\in Y\}$</code> is a finite union of arithmetic progressions. Here $f:X\to X$ is a morphism of a (smooth) projective variety, and $Y\subset X$ is a (smooth) subvariety. http://mathoverflow.net/questions/128251/a-counterexample-of-the-birch-and-swinnerton-dyer-conjecture Comment by Joe Silverman Joe Silverman 2013-04-21T14:54:54Z 2013-04-21T14:54:54Z No. (If there were one, it would be big news, you could find it by googling &quot;Birch and Swinnerton-Dyer counterexample&quot;.) http://mathoverflow.net/questions/128176/will-quantum-computing-kill-cryptography/128179#128179 Comment by Joe Silverman Joe Silverman 2013-04-20T19:09:34Z 2013-04-20T19:09:34Z Quantum algorithms for CVP and SVP have been studied since Shor's original paper, which points to them as interesting problems to study. But I would never try to argue that lattice-based cryptosystems are safe simply because some people have spent some time trying to break them. All that one can currently honestly say about <i>any</i> practical cryptosystem is that as of today, no one has publicly given an algorithm that breaks it. (Notice both the phrase &quot;as of today&quot; and the word &quot;publicly&quot; in that last sentence!) http://mathoverflow.net/questions/128176/will-quantum-computing-kill-cryptography/128179#128179 Comment by Joe Silverman Joe Silverman 2013-04-20T18:28:06Z 2013-04-20T18:28:06Z Ummmm... Possibly, I'm not sure what the &quot;dihedral hidden subgroup problem&quot; is. But one might as easily say &quot;a polynomial time quantum algorithm to solve CVP would break these cryptosystems.&quot; My point was that at the moment, we don't know a quantum algorithm that would break these lattice-based cryptosystems in subexponential time (much less polynomial time). Might such algorithms exist? Sure. But lacking lower bounds in complexity theory, all we can do is talk about the best <i>known</i> algorithms. OTOH, for factoring, DLP, and ECDLP, we already have poly time quantum algorithms. http://mathoverflow.net/questions/126554/degree-of-field-extension Comment by Joe Silverman Joe Silverman 2013-04-04T20:30:27Z 2013-04-04T20:30:27Z However, if you meant to write $\mathbb{Q}(\xi)$, then this is a standard result in an introductory undergraduate course in number theory and/or field theory, so it really is not a suitable question for this site. Please read the site FAQ. You might try posting your question on Math Stackexchange. The same question when $p$ is composite is also interesting, but more difficult. http://mathoverflow.net/questions/125664/division-by-3-on-elliptic-curve/125848#125848 Comment by Joe Silverman Joe Silverman 2013-03-30T17:40:53Z 2013-03-30T17:40:53Z @Michael: Sorry, you're right. Given a specific point, its divisibility by 3 (or by 2, or by $m$) is a purely algebraic question. However, usually the reason to do this is to compute the Mordell-Weil group, so (for $m=2$) one looks at equations $x-\alpha_i=y_i^3$ for $i=1,2,3$. And then the arithmetic of $L$ will come into play. For $X_1(11)$ and $m=2$, my recollection is that $L$ has class number 1 and that one can completely characterize the units that are squares, which was crucial in showing that $X_1(11)(\mathbb{Q})$ has rank $0$. http://mathoverflow.net/questions/125664/division-by-3-on-elliptic-curve/125848#125848 Comment by Joe Silverman Joe Silverman 2013-03-29T12:45:57Z 2013-03-29T12:45:57Z @Michael: I had assumed the OP was primarily interested in the situation where one can reduce to questions about cubes in $K$, but you're quite right that this is the right generalization if one doesn't assume that the 3-torsion is rational. Of course, now the arithmetic of $L$ will come into play, especially the class group and unit group. I first saw this used (for 2-torsion) in a letter that Tate wrote computed $E(Q)$ for $E=X_1(11)$. For general $m$, Bob Wake in his thesis partially generalized the idea of looking at $L^*/{L^*}^m$, but I don't think that he ever published his thesis. http://mathoverflow.net/questions/125664/division-by-3-on-elliptic-curve/125668#125668 Comment by Joe Silverman Joe Silverman 2013-03-28T21:06:33Z 2013-03-28T21:06:33Z @Enrique: I didn't say it was explicit, I said it's where the result comes from. In principle, one should be able to make it explicit by writing down explicit functions F, G and H, but first you'd need to write the elliptic curve in a form where its $m$-torsion is rational. So there's a fair amount of work involved in doing this, and I don't recall seeing it. But at least for $m=3$, it's likely to be feasible.