1
$\begingroup$

I came across this sentence in one of the material I was reading this below :

A major result is the Hasse-Minkowski Principle, which implies that a curve C has a point over $\mathbb{Q}$ iff it has a point over $\mathbb{R}$ and over every local field $\mathbb{Q}_p$. This also implies the points of genus zero curve over $\mathbb{Q}$ can all be determined easily.

But the local solubility criteria says nothing about how one actually finds a global solution. Can someone clarify this for me.

$\endgroup$
  • 1
    $\begingroup$ Over $\mathbb Q_p$, solutions are not so hard to find because this is easily reduced (a la Hensel's lemma) to finding them over the finite rings $\mathbb Z/p^n\mathbb Z$ and lifting them to $\mathbb Q_p$. Then by comparing a (finite) number of values of $p$ (often including $p=\infty$), we can reduce this to a finite number of candidates. $\endgroup$ – RKD May 25 '17 at 18:31
0
$\begingroup$

You are correct in that the local-to-global principle only let's you decide whether there are global solutions, but it does not offer a way to easily find the rational points. For instance, let $C$ be the Pell equation $$C: x^2-109y^2=-1.$$ Then, it is easy to use the Hasse-Minkowski theorem to deduce that there must be global solutions (you only need to check there are solutions over $\mathbb{R}$ and over $\mathbb{Q}_{109}$, and by Hensel's lemma, you just need to check for solutions in $\mathbb{Z}/109\mathbb{Z}$ for the latter). However, the simplest solution is given by $$x=8890182, \text{ and } y = 851525.$$ The local-to-global solution does not tell you how to find it. There are "elementary methods" to find such a solution (e.g., continued fractions), but what elementary means is on the eye of the beholder.

$\endgroup$
  • $\begingroup$ thank you for explaining this to me. $\endgroup$ – shahrina ismail Jun 14 '17 at 2:30
  • $\begingroup$ You are welcome! $\endgroup$ – Álvaro Lozano-Robledo Jun 14 '17 at 17:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.