Joe Silverman
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Registered User
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Professor of Mathematics, Brown University.
Primary interests: number theory, arithmetic geometry, elliptic curves, arithmetic dynamics, cryptography.
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1d |
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Help with this system of Diophantine equations 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 "intrinsic" 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. |
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May 16 |
accepted | Ramification in Division field of Abelian Varieties |
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May 16 |
answered | Ramification in Division field of Abelian Varieties |
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May 8 |
accepted | Short basis for the unit group of a number field |
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May 4 |
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Short basis for the unit group of a number field @Asaf: Aren't real quadratic fields the 1-dimensional case? So your comments are probably more relevant to, say, a totally real cubic field. |
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May 4 |
answered | Short basis for the unit group of a number field |
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May 4 |
awarded | ● Nice Answer |
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May 3 |
accepted | elliptic curve with a degree 2 isogeny to itself? |
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May 3 |
answered | elliptic curve with a degree 2 isogeny to itself? |
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May 2 |
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Torsion subgroups in families of twists of elliptic curves 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 $$ \operatorname{rank} E_d(\mathbb{Q}) + \operatorname{rank} E(\mathbb{Q}) \ = \operatorname{rank} E(\mathbb{Q}(\sqrt{d})) $$ |
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Apr 26 |
answered | Integer dynamics hitting infinitely many primes |
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Apr 26 |
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Integer dynamics hitting infinitely many primes @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 $\{n : f^n(x)\in Y\}$ 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. |
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Apr 20 |
awarded | ● Enlightened |
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Apr 20 |
accepted | Will quantum computing kill cryptography ? |
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Apr 20 |
awarded | ● Nice Answer |
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Apr 20 |
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Will quantum computing kill cryptography ? 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 any practical cryptosystem is that as of today, no one has publicly given an algorithm that breaks it. (Notice both the phrase "as of today" and the word "publicly" in that last sentence!) |
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Apr 20 |
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Will quantum computing kill cryptography ? Ummmm... Possibly, I'm not sure what the "dihedral hidden subgroup problem" is. But one might as easily say "a polynomial time quantum algorithm to solve CVP would break these cryptosystems." 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 known algorithms. OTOH, for factoring, DLP, and ECDLP, we already have poly time quantum algorithms. |
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Apr 20 |
answered | Will quantum computing kill cryptography ? |
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Mar 30 |
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Division by 3 on elliptic curve @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 $0$. |
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Mar 29 |
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Division by 3 on elliptic curve @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. |
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Mar 28 |
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Division by 3 on elliptic curve @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. |
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Mar 26 |
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Division by 3 on elliptic curve @wccanard: That was my first thought, too. Except it's not quite right, because if one of the three quantities is 0, then you really do need to check that the other two are squares. So my answer isn't quite right, either. It will work as long as P is not itself an m-torsion point, but if F(P) or G(P) vanishes, then there will be a third condition H(P) that needs to be checked. I believe that the ambiguity comes from making a specific choice of an isomorphism $E[m]\cong\mathbb{Z}/m\mathbb{Z}\times\mathbb{Z}/m\mathbb{Z}$. |
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Mar 26 |
answered | Division by 3 on elliptic curve |
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Mar 8 |
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degree of sum algebraic functions @Martin: The degree of a rational function $f$ on a curve $C$ is the degree of the induced map $f:C\to\mathbb{P}^1$. (Constant maps are assigned degree 0.) Alternatively, it's the degree of the polar part of div$(f)$. Of course, one can't add functions that map to $\mathbb{P}^1$, since $\mathbb{P}^1$ is not a group. But one can treat $f$ as a rational map $f:C\dashrightarrow\mathbb{A}^1$, and then it makes sense to add two functions. In any case, this seems more of an exercise than a research-level question, so probably better suited to mathstackexchange. |
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Mar 7 |
revised |
The elliptic Lehmer problem for several independent algebraic points Added further information about the problem |
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Mar 6 |
accepted | The elliptic Lehmer problem for several independent algebraic points |
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Mar 6 |
answered | The elliptic Lehmer problem for several independent algebraic points |
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Mar 4 |
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Which level structures on elliptic curves are twist-invariant? @Jordan: Is that still true if there are twists by 4th or 6th roots of unity? Or are you just considering quadratic twists? |
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Mar 4 |
awarded | ● Nice Answer |
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Mar 1 |
answered | Effective Mordell |
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Feb 26 |
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Genus of Y^3 = X^4 - 1. Hmmmm.... I suspect that the OP wanted to know the genus of a smooth projective model, which would be the designularization $\hat C$ of the projective curve given in homogeneous coordinates by $C:X^n+Y^mZ^{n-m}=Z^n$. (For concreteness, I've assumed that $n\ge m$.) The genus of $\hat C$ will not, in general, be $(n-1)(m-1)/2$. I don't recall offhand the formula, but my recollection is that it involves $\gcd(m,n)$, and as I indicated in an earlier comment, is computable by a short exercise in blowing up. |
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Feb 26 |
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Genus of Y^3 = X^4 - 1. Possibly "teach me" means "explain how to compute", rather than "tell me the answer". One can check that a curve of this sort is nonsingular (as a projective curve) in a minute or two by hand, one really doesn't need Magma. Indeed, it's a nice exercise in a first-year algebraic geometry course to compute the genus of X^n + Y^m = 1 (by hand!). The sequence of blow-ups needed to resolve the singularity mimics the Euclidean algorithm used to compute gcd(m,n). |
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Feb 17 |
awarded | ● Enlightened |
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Feb 17 |
awarded | ● Nice Answer |
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Feb 9 |
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What kind of subset is Spec(R_P) in Spec(R)? Isn't Spec$(R_P)$ (considered as a subset of Spec$(R)$) the complement of $V(P)$? That would make it open. (This question seems a bit elementary for MO.) |
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Feb 3 |
answered | Trichotomies in mathematics |
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Jan 31 |
accepted | The height of an orbit under rational self-maps |
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Jan 31 |
answered | The height of an orbit under rational self-maps |
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Jan 24 |
accepted | Reference request for the theory of heights over function fields |
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Jan 23 |
awarded | ● Necromancer |
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Jan 22 |
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modern reference for Néron’s “Quasi-fonctions et Hauteurs sur les Varietes Abeliennes” Isn't this pretty much the content of Lang's Fundamentals of Diophantine Geometry, Chapter 11, "Neron Functions on Abelian Varieties"? |
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Jan 21 |
answered | Reference request for the theory of heights over function fields |
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Jan 5 |
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Can the difference of two distinct Fibonacci numbers be a square infinitely often? > Does some generalization of the abc conjecture predict something? As was shown in one (or more) of the answers, the solutions lead to integer points on an affine piece of one of a few K3 surfaces. Vojta's conjecture implies that the set of integer points on an affine K3 surface lie on a finite union of curves. One can certainly view Vojta's conjecture as a generalization of the ABC conjecture, since it implies ABC. So this is a possible answer to joro's "Added much later" question. |
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Jan 4 |
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Can the difference of two distinct Fibonacci numbers be a square infinitely often? AFAIK, there are no proven major results on integer points on affine pieces of K3 surfaces. However, Vojta's conjecture predicts that the set of such points lies on a finite union of curves. So assuming Vojta's conjecture, one might be able to make further progress. |
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Dec 30 |
awarded | ● Yearling |
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Dec 29 |
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Old books still used @ayanta: Well, the new chapter on elliptic curves was written with an eye towards fitting into the style of the rest of the text. (An assertion that I feel that I'm entitled to state as a fact, rather than as an opinion.) So I guess there might be some who would say that the elliptic curves chapter is also "outdated", despite having been written quite recently! But I have to respectively disagree with your opinion of the book, which I feel is a masterpiece. |
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Dec 29 |
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Old books still used @Qfwfq: Well, we used it when I was a junior, so it had already appeared in 1975. But I don't know the original publication date offhand. |
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Dec 26 |
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Arithmetic dynamics and dynamics on moduli spaces A quick answer to (2) is that there are a number of conjectures that are analogues from arithmetic geometry that are helping to drive the field of arithmetic dynamics. These include dynamical analogues of (1) the uniform boundedness conjecture for torsion on abelian varieties; (2) Raynaud's theorem on torsion points on subvarieties of abelian varieties; (3) Faltings' theorem (Mordell-Lang conjecture) on rational points on subvarieties of abelian varieties. Replace torsion points with preperiodic points and Mordell-Weil groups with orbits to get rough analogues, although there are subtleties. |
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Dec 26 |
answered | Arithmetic dynamics and dynamics on moduli spaces |
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Dec 24 |
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If rational points are like entire curves, then what do algebraic points correspond to @Masse: Actually, for quadratic points, gonality is sharp. In other words, if X is not hyperelliptic, then it has only finitely many quadratic points. My recollection is that the same is true for cubics (but I may be misremembering), and in any case, it's not true in general for higher degree. For references, there are various papers by Joe Harris, Dan Abramovich, and me. (I think that the first one was mine and Joe Harris's that did the quadratic points case.) I'm not quite sure which result of Faltings-Frey you mean, but of course all these results rely on Faltings' Annals paper. |
