most recent 30 from http://mathoverflow.net 2013-05-18T22:04:23Z http://mathoverflow.net/feeds/user/949 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/107960/how-to-compute-the-picard-rank-of-a-k3-surface/107963#107963 Ari Shnidman 2012-09-24T07:52:32Z 2012-09-24T07:52:32Z

<p>There are some papers of <a href="http://www.math.leidenuniv.nl/~rvl/papers.html" rel="nofollow">van Luijk</a>, where he computes the ranks of some K3s over number fields. The trick is to note that $NS(X) \hookrightarrow NS(X_p)$, where $X_p$ is the reduction of $X$ modulo a prime ideal $p$. One can determine the rank of $NS(X_p)$ by counting eigenvalues of Frobenius which differ from $q$ (the size of the residue field) by a root of unity. If you want to find rank 1 K3s, you can reduce modulo two different primes and hope to find rank 2 reductions which have lattices which are incompatible in some sense, forcing $NS(X)$ to be rank 1. (The issue here is that the rank of $NS(X_p)$ will always be even, so you can't win by using a single prime.)</p> <p>I'm not sure how this works when you want to find K3s of larger rank though, unless you had a way of exhibiting linearly independent divisor classes. Anyhow, van Luijk uses this technique to find rank 1 quartics in $\mathbb{P}^3$ and I think others have done the same with genus 2 K3s defined over $\mathbb{Q}$. </p> <p>I should add that the situation is much easier for Kummer surfaces. If I'm not mistaken, the rank of $X = K(A)$ ($A$ is an abelian surface) is 16 plus the Picard rank of $A$. The 16 comes from the 16 exceptional divisors you get when you blow up $A$ at its 2-torsion points. The rank of $A$ is usually not hard to figure out: a generic $A$ has rank 1, if $A$ is a product of elliptic curves then its rank is 2,3 or 4 depending on whether the curves are isogenous and whether they have CM or not, and there are a few other cases which one can probably figure out... </p>

http://mathoverflow.net/questions/3448/are-there-two-non-isomorphic-number-fields-with-the-same-degree-class-number-and Ari Shnidman 2009-10-30T16:04:22Z 2012-09-17T19:29:51Z

<p>If so, do people expect certain invariants (regulator, # of complex embeddings, etc) to fully 'discriminate' between number fields?</p>

http://mathoverflow.net/questions/107301/isogeny-classes-of-elliptic-curves/107325#107325 Ari Shnidman 2012-09-16T15:39:21Z 2012-09-16T15:39:21Z

<p>One way to get a counterexample is to take any elliptic curve $E$ over a quadratic field $K$ whose conductor is a prime ideal $\mathfrak{p}$ lying over a split prime $p$ in $\mathbb{Q}$. This means that $E$ has multiplicative reduction at $\mathfrak{p}$. If $\sigma$ is the non-trivial automorphism of $K/\mathbb{Q}$, then $E^\sigma$ has conductor $\mathfrak{p}^\sigma \neq \mathfrak{p}$, so $E^\sigma$ has good reduction at $\mathfrak{p}$. Any isogeny $E \to E^\sigma$ would be defined over a number field, and multiplicative/good reduction are both stable under finite field extension. Since isogenous curves have the same reduction type, $E$ and $E^\sigma$ cannot be isogenous.</p> <p>An example: the two curves of minimal norm conductor over $\mathbb{Q}(\sqrt{5})$ have norm conductor 31, but their conductors are the two different ideals of norm 31. Check out William Stein's table: <a href="http://modular.math.washington.edu/Tables/hmf/sqrt5/ellcurve_aplists.txt" rel="nofollow">http://modular.math.washington.edu/Tables/hmf/sqrt5/ellcurve_aplists.txt</a> </p> <p>As a double check, you can even tell from the Hecke eigenvalues in the table that the curves are conjugate to one another.</p> <p>Maybe someone can give a simple counterexample (without using Elkies' theorem quoted by stankewicz) with integral $j$-invariant?</p>

http://mathoverflow.net/questions/104945/on-the-jacobian-origin-of-cm-abelian-varieties/104966#104966 Ari Shnidman 2012-08-18T03:08:59Z 2012-08-18T06:10:07Z

<p>When $n &lt; 4$, $A$ is isogenous to the Jacobian of a stable curve because the Torelli locus is dense. This should imply that $A$ is isogenous to the Jacobian of a smooth curve because $A$ is simple (at least if the CM-type is primitive). </p> <p>If $n \geq 4$, then conditional results of Chai-Oort (see their recent Annals paper) imply that $A$ is not necessarily isogenous to the Jacobian of a curve. Tsimerman then proved this unconditionally. I would guess that the set of CM points whose isogeny orbit is disjoint from the Torelli locus is dense in $\mathcal{A_g}$, but I don't know. </p>

http://mathoverflow.net/questions/19886/decomposition-of-primes-where-the-residue-field-extensions-are-allowed-to-be-ins/19893#19893 Ari Shnidman 2010-03-30T23:25:51Z 2010-03-30T23:25:51Z

<p>I believe that's right, at least when $S$ is finitely generated over $R$. See Serre's Local Fields page 21-22 (in the English translation); he states his assumptions on page 13. </p>

http://mathoverflow.net/questions/5351/whats-an-example-of-a-space-that-needs-the-hahn-banach-theorem/5400#5400 Ari Shnidman 2009-11-13T17:46:18Z 2009-11-13T17:46:18Z

<p>I don't fully understand what you're asking, but the one proof I know of the fact that (up to topological isomorphism) the only complete Archimedean fields are $\mathbb{C}$ and $\mathbb{R}$ uses the Hahn-Banach theorem.</p>

http://mathoverflow.net/questions/5283/are-there-generalisations-of-a-limit-for-just-divergent-sequences/5304#5304 Ari Shnidman 2009-11-13T02:56:34Z 2009-11-13T02:56:34Z

<p>Cesaro summation (the process which you describe) defines a linear functional on a subspace of the Banach space of bounded sequences (namely those sequences which are cesaro summable). Using Hahn-Banach (or one of its variants), one can extend this linear functional to the whole space of bounded sequences, and the extension WILL be shift invariant. However, the extension is not unique and existence depends on the Axiom of choice. </p> <p>See the Wikipedia entry for <a href="http://en.wikipedia.org/wiki/Banach%5Flimit" rel="nofollow">Banach limit</a> for more info. </p>

http://mathoverflow.net/questions/3591/mathematicians-who-were-late-learners-list/3607#3607 Ari Shnidman 2009-10-31T21:51:36Z 2009-10-31T21:51:36Z

<p>Dwork started out as an electrical engineer and was 31 when he received his PhD. The memorial article by Tate and Katz gives the interesting details. </p>

http://mathoverflow.net/questions/117296/characterizing-primes-that-split-completely-vs-primes-with-a-given-splitting-beh Ari Shnidman 2012-12-27T02:45:43Z 2012-12-27T02:45:43Z

Your second sentence is not true, unless $L/K$ is Galois.

http://mathoverflow.net/questions/104945/on-the-jacobian-origin-of-cm-abelian-varieties/104966#104966 Ari Shnidman 2012-08-18T16:53:26Z 2012-08-18T16:53:26Z

Sorry, I was a bit terse. The open Torelli locus (i.e. Jacobians of smooth curves) is dense in $\mathcal{A}_n$ (the moduli space of $n$-dimensional ppav) for $n &lt; 4$, so the closure is the whole thing. But it's known that anything in (closure minus open locus) is a product of Jacobians (see Mumford's Woods Hole notes). So if $A$ is simple then it's isogenous to a Jacobian.

http://mathoverflow.net/questions/95121/elliptic-curves-and-torsion-points Ari Shnidman 2012-04-25T07:57:14Z 2012-04-25T07:57:14Z

x^3 + y^3 = p, with p &gt; 3 a prime

http://mathoverflow.net/questions/3448/are-there-two-non-isomorphic-number-fields-with-the-same-degree-class-number-and/29467#29467 Ari Shnidman 2012-03-16T06:10:03Z 2012-03-16T06:10:03Z

There are examples with smaller absolute discriminant: the &quot;pure&quot; cubic fields obtained by adjoining cube roots of 6 and 12. Both have discriminant -972 and class number 1, but they're not isomorphic. Pure cubics give lots of examples like this -- I was able to find 5 fields with common discriminant and class number.