User amateur algebraist - MathOverflowmost recent 30 from http://mathoverflow.net2013-05-18T12:03:39Zhttp://mathoverflow.net/feeds/user/2120http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/8219/elementary-theory-of-finite-fieldsElementary theory of finite fieldsamateur algebraist2009-12-08T17:54:12Z2009-12-09T17:15:39Z
<p>I read on Ax's article that the elementary theory of finite fields is decidable if one assumes the continuum hypothesis to be true. What about if one assumes the hypothesis to be false?</p>
http://mathoverflow.net/questions/7907/how-to-find-all-integer-points-on-an-elliptic-curveHow to find all integer points on an elliptic curve?amateur algebraist2009-12-05T22:09:21Z2009-12-07T17:42:13Z
<p>How can I determine the integer points of a given elliptic curve if I know its rank and its torsion group? </p>
<p>I read same basic books on elliptic curves but as a non-professional I didn't understand everything. Is it true that if rank is 0 and torsion group is isomorphic to a group of order $n$ then the number of integer points is $n-1$? And what is a good reference to learn to determine the integer points if the rank is positive? </p>
<p>I tried to read the book Rational Points on Elliptic Curves but I didn't found an explicit algorithm. I just heard something like take some point and use group law to find the rest. But how can I be sure that I have found every point?</p>
<p>The curve I had on my mind is $2x^3 + 385x^2 + 256x - 58195 = 3y^2$. I'm not even sure if this is an elliptic curve. I mean why it is projective and why it is isomorphic to a closed subvariety of $\mathbb{P}_{\mathbb{Q}}^2$? And why it contain the priviledged rational point $(0,1,0)$? </p>
http://mathoverflow.net/questions/3699/minimizing-a-functional/6976#6976Answer by amateur algebraist for Minimizing a functionalamateur algebraist2009-11-27T19:25:51Z2009-11-27T19:25:51Z<p>Sorry. The problem is that we start from the point $(0,0)$ and we swim to $(1,0)$. Now if you swim by straight line you won't necessary end to the point where you want to. Then you have to compute how the flow affects your swimming speed and the direction you are swimming to. I think this leads to the functional equation I got if I computed it correctly. Then you can probably find a solution curve which tells the direction you have to swim with respect to time.</p>
http://mathoverflow.net/questions/8219/elementary-theory-of-finite-fieldsComment by amateur algebraistamateur algebraist2009-12-08T20:05:39Z2009-12-08T20:05:39ZAx, James The elementary theory of finite fields. Ann. of Math. (2) 88 1968 239–271. http://mathoverflow.net/questions/8219/elementary-theory-of-finite-fields/8234#8234Comment by amateur algebraistamateur algebraist2009-12-08T20:05:04Z2009-12-08T20:05:04ZI was wondering whether there is a ring isomopphism $\prod_p\mathbb{F}_p/\bigoplus \mathbb{F}_p\approx \prod_p \mathbb{F}_{p^p}/\bigoplus \mathbb{F}_{p^p}$ if we don't assume continuum hypothesis. Probably this theorem or is not mandatory to prove the decidability result even if CH is assumed. But it would be nice to know why this does/doesn't hold if CH is not assumed.http://mathoverflow.net/questions/7907/how-to-find-all-integer-points-on-an-elliptic-curve/7979#7979Comment by amateur algebraistamateur algebraist2009-12-06T15:32:52Z2009-12-06T15:32:52ZOkay. But how can I prove those are the only one? Am I right that the priviledged rational point is not an integer point?http://mathoverflow.net/questions/7907/how-to-find-all-integer-points-on-an-elliptic-curve/7912#7912Comment by amateur algebraistamateur algebraist2009-12-05T23:57:49Z2009-12-05T23:57:49ZUnfortunately the curve I had on my mind has larger conductor than 130000http://mathoverflow.net/questions/7907/how-to-find-all-integer-points-on-an-elliptic-curveComment by amateur algebraistamateur algebraist2009-12-05T22:26:51Z2009-12-05T22:26:51ZThanks! I have one curve which is of rank 4 and torsion subgroup isomorphic to trivial abelian group so I would like to know some method to prove the solutions I found are the only one.