A common definition of an elliptic curve over a field $k$, is that it is a smooth projective curve of genus $1$ (defined over $k$) with a distinguished $k$-rational point. The distinguished point is particularly important in the definition, as not all smooth projective curves of genus 1 will be elliptic curves. For instance, $y^2=-x^4-1$ will not be an elliptic curve over $\mathbb{Q}$.

However, I've heard that there are indeed some fields where every smooth projective curve of genus 1 will be elliptic; in particular that this is true for non-archimedean local fields with good reduction. How can one prove that there's a distinguished point for this field?


Good reduction is not a property of the field, but of the curve.

What you mean to write is: if $C$ is a genus one smooth proj. curve over a non-arch. local field with good reduction then $C$ admits a rational point.

The proof is as follows: by Hensel's lemma (or Newton's method, or whatever you like to call it), it is enough to check the corresponding statement for the reduction of $C$ over the residue field.

But over a finite field, any genus $1$ curve has a rational point, by the Weil bounds. (If the res. field has order $q$, the number of points is at least $1 + q - 2 \sqrt{q}$, which is always positive.)

Note, though, that there will typically be more than one rational point over the residue field, and certainly there will be more than one rational point over the local field. None of them is distiguished; you still have to choose one if you want to make your genus $1$ curve an elliptic curve.

  • $\begingroup$ Oh I see. I had considered it for a finite field under a more obscure argument, since I wasn't aware you could apply the Weil bounds to any such curve over a finite field (i.e. only for an elliptic curve, but a quick recheck shows that isn't true). Hensel's lemma easily does the trick for lifting our solution. $\endgroup$ – potionboy May 9 '13 at 5:57

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.