Joe Silverman
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 49m answered Finite orbits on an elliptic curve with two generic involutions 20h comment Diophantine equations over natural numbers If you have a general method to solve some reasonable set of equations in $k$ variables using positive integers, then simply replacing subsets of the variables by their negatives, you'll get $2^k$ equations, and solving each of them in positive integers gives the full set of solutions to the original equation in arbitrary integers. So you probably need to be more specific about the problem that you're studying, since there may well be cases where one can find all positive solutions, but it's hard to find all integer solutions. Oct 1 comment A characteristic 2 polynomial recursion Hi Paul, Welcome to MathOverflow. I edited your question to make it easier to read, hope you don't mind. I also fixed the Z_2 by changing it back to Z/2Z. Interesting question. Oct 1 revised A characteristic 2 polynomial recursion Fixed the LaTeX to make it easier to read Sep 29 answered Most intriguing mathematical epigraphs Sep 25 comment Elliptic curve model minimal This was asked yesterday mathoverflow.net/questions/219049/elliptic-curve-minimal-model, put on hold, and also, Vesselin Dimitrov gave a reference for the answer in the comments. Please don't re-post the same question, at least not until you've looked at the reference that was provided. Sep 24 comment Do there exist elliptic curves over schemes which have all primes as residue characteristics? That's a fancy way to do it. :) The fact that elliptic curves with integral $j$-invariant have everywhere good reduction over an extension field (which is iff) follows easily if char $\mathfrak p\ge5$ using the Weierstrass eqn, and can even be done for 2 and 3 with more work. There's also the Serre-Tate (Neron-Ogg-Shafarevich) criterion which says good reduction iff the action of Galois on appropriate torsion is unramified. So adjoining some torsion gives a field of everywhere good reduction. In particular, I think you can take your $L$ to be something like $K(E[15])$. Sep 24 awarded Nice Answer Sep 24 answered Do there exist elliptic curves over schemes which have all primes as residue characteristics? Sep 22 comment Teaching an abstract algebra class involving modules, best way to introduce operations on modules? @roysmith The functorial definition provides at least a partial answer. Given a bunch of $M_i$, it is natural to want to look at a module that naturally "contains" all of the $M_i$ and is as small as possible (that's the direct sum) and a module that naturally maps to all of the $M_i$ and is as small as possible (that's the direct product). Of course, one can also motivate from the students knowledge that if $K$ is a field, then $K^n$ is very interesting, and it's not unnatural to ask what happens if we allow $n$ to "be infinite". Sep 22 comment Teaching an abstract algebra class involving modules, best way to introduce operations on modules? A more modern treatment is that the direct sum and direct product of $\{M_i\}_{i\in I}$ is a module $M$ with maps either from or to the $M_i$ that is largest or smallest, in the sense that if there is any other module $N$ with such maps, then there is a unique map to or from $M$ commuting with the given maps. From this, one can show that your definitions prove that direct sums and products exist in the category of modules, and they are unique (up to unique isomorphism) by functoriality. Also, the product you define should be the set of maps $m:I\to M$ such that $m_i\in M_i$. Sep 22 comment Geometric / physical / probabilistic interpretations of Riemann zeta(n>1)? @DavidRoberts It seems to come from 10=4+6, since $A$ has weight 4 and $B$ has weight 6. But I must admit that I expected to see $12=\text{LCM}(4,6)$, as happens more commonly with modular form. Sep 22 answered Geometric / physical / probabilistic interpretations of Riemann zeta(n>1)? Sep 20 comment On discriminants of elliptic curves Did you mean to put some conditions on $m_1$ and $m_2$. Certainly you don't want $m_1=m_2$. But actually, if $d=\gcd(m_1,m_2)$, then your intersection always trivially contains $\mathbb Q(E[d])$. So the interesting case is probably when $\gcd(m_1,m_2)=1$. Also your proof of your example needs to note that if the 2-torsion is rational, then you want to take $n=1$, not $n=2$. Sep 19 comment What should I read to prepare for research in Number Theoretic Cryptography? Maybe try MathStackExchange Cryptography. But as Chris said, a good source is your supervisor, or if you don't have one yet, then go talk to all of the faculty at your university who do research in cryptography, since your eventual supervisor will undoubtedly be chosen from that set. Sep 18 comment Why are most coefficients of these minimal polynomials divisible by $p$? Good point. I did say that my argument was for $a$ being an algebraic integer (or at least, being $p$-adically integral). As you note, if one puts a denominator in $a$ that exactly kills the $p$-adic valuation of $1+\zeta$, the conclusion fails. Nice example! Sep 18 comment Why are most coefficients of these minimal polynomials divisible by $p$? @Wolfgang Let $u$ and $v$ be algebraic numbers. Let $u_1,\ldots,u_n$ be the Galois conjugates of $u$, and let $v_1,\ldots,v_m$ be the Galois conjugates of $v$. Then $\{u_iv_j : 1\le i\le n,1\le j\le m\}$ is a complete set of Galois conjugates of $uv$, although there may, of course, be repeated values. This is clear, since $\sigma(uv)=\sigma(u)\sigma(v)$ for any $\sigma\in\text{Gal}(\overline{\mathbb Q}/\mathbb Q)$. So in the OP's case, the conjugates of $(1-\zeta^n)a$ all have the form $(1-\zeta^{nj})b$, where $b$ is some Galois conjugate of $a$. Sep 17 comment Why are most coefficients of these minimal polynomials divisible by $p$? @GerhardPaseman I guess I'm just too used to taking $\pi$ to be 1 minus a $p$'th root of unity, with the norm of $\pi$ generating the ideal $p\mathbb Z$. But you're probably right that for this problem it would be easier to just use $1+\zeta$ with $\zeta$ a primitive $2p$'th root of unity. Sep 17 revised Why are most coefficients of these minimal polynomials divisible by $p$? added 89 characters in body Sep 17 answered Why are most coefficients of these minimal polynomials divisible by $p$?