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Every elliptic curve over a field $K$ can be mapped to a smooth, projective genus 1 curve (also defined over $K$) with a $K$-rational point, and vice versa. As I understand it, curves without said $K$-rational points can never be mapped to elliptic curves (since clearly then the elliptic curve group law cannot be established).

I'm trying to think of an example of an affine curve whose genus is $1$, but which is not also an elliptic curve. I know that the Riemann-Roch theorem gives:

$$g = \frac{(d-1)(d-2)}{2} - s$$

Some searching online yielded the curve $y^{2}=x^{5}-x$, however I'm relatively certain that this Riemann-Roch gives $g = 6$ in $\mathbb{P}^2$, not $1$.

What are some examples of affine genus $1$ curves that are not also elliptic curves?

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    $\begingroup$ As a note, the curve $y^2 = x^5 - x$ has genus $2$, not $6$. This can be shown by applying the Riemann-Hurwitz Theorem to the projection map $(x,y) \mapsto x$. $\endgroup$ Commented Mar 6 at 17:08
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    $\begingroup$ What is your definition of "elliptic curve"? $\endgroup$ Commented Mar 7 at 4:47

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Consider an equation of the form $y^2=f(x)=a_0x^4+a_1x^3+\cdots+a_4$, where $a_i\in \mathbb{Q}$. If the discriminant of the quartic is non-zero, then this defines a smooth curve of genus $1$. However, you have to be careful with what that means: if you projectivise this naively, i.e. you treat this as a curve in $\mathbb{P}^2$, then it is easy to check that the point at infinity will be singular.

On the other hand, it is a general fact that there is a unique (up to isomorphism) smooth curve whose function field is $\mathbb{Q}(x,y)/(y^2-f(x))$. It's just that the curve does not embed as a smooth curve in $\mathbb{P}^2$. It does, however, embed as a smooth curve in $\mathbb{P}^3$ via $[X_0,X_1,X_2,X_3] = [1,x,y,x^2]$. It turns out that this projective curve has the affine piece $X_0\neq 0$ described by the above equation, and the only missing points are $[0,0,\pm\sqrt{a_0},1]$. This is explained in detail in Silverman, The Arithmetic of Elliptic Curves, Chapter II, Example 2.5.1.

Now, if $a_0$ is not a square in $\mathbb{Q}$, then those two points at infinity will not be rational points; and it is easy to arrange $f(x)$ in such a way that the affine piece will also not have any rational points, e.g. by making sure that it has no points over some completion of $\mathbb{Q}$. A very simple example: take $f(x) = -x^4 - 1$. Since $-1$ is not a rational square, the points at infinity, $[0,0,\pm\sqrt{-1},1]$, are not defined over $\mathbb{Q}$; and the affine piece certainly has no rational points, since it does not even have real points: $f(x)$ is negative for all $x$, so cannot be equal to $y^2$ for any real $y$.

The two examples in Pont's answer are more complicated: they are "notorious" for having points over all completions of $\mathbb{Q}$, but none over $\mathbb{Q}$, which is then necessarily harder to prove.

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The following are classical examples of genus $1$ curves that are not elliptic curves over $\Bbb{Q}$:

$1.$ $C : 3X^3 + 4Y^3 + 5Z^3 = 0$ in $\mathbb{P}^2_{\mathbb{Q}}$ (Selmer)

$2.$ Compactification of an affine curve $2y^2 = x^4 - 17$ in $\mathbb{P}^3_{\mathbb{Q}}$(Lind)

You can prove that these curves have genus $1$, but they do not have $\mathbb{Q}$-rational points.

I will explain each of the curves in more detail.

$1.$ Indeed, $3X^3+4Y^3+5Z^3=0$ is a torsor of an elliptic curve $E: X^3+Y^3+60Z^3=0$(This has a rational point $[1:-1:0]$).You can prove $C(\Bbb{Q})=\emptyset$ by showing that the rank of $E/\Bbb{Q}$ is $0$.This kind of discussion is detailed in Cassels''Lectures on elliptic curves'.

$2.$ Note that projective closure of this affine curve in $\mathbb{P}^2_{\mathbb{Q}}$ has singular point at infinity $[0:1:0]$, thus we take projective closure in $\mathbb{P}^3_{\mathbb{Q}}$ not in$\mathbb{P}^2_{\mathbb{Q}}$.The point at infinity contains $\sqrt{2}$ in its coordinate, so there are no $\Bbb{Q}$-rational point at infinity. Also, you can check elementary the affine curve does not have $\Bbb{Q}$ rational points(cf.Chapter Ⅹ of Silverman's 'The arithmetic of elliptic curves'). You can also use Brauer-Manin obstruction to prove the affine curve does not have $\Bbb{Q}$-rational points by computing Hilbert symbol.

The equation $y^2 = -x^4 - 1$ in Alex B.'s answer provides a simpler example in response to this question. The two examples in my answer illustrate the so-called counterexample to local-global principle for genus $1$ curves.

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  • $\begingroup$ @Piquito, what is your definition of an elliptic curve? The definition I am using is that of a genus 1 curve with a base point (see Silverman's book). The crucial point is that the curve defined by $3x^3 + 4y^3 + 5z^3 = 0$ is not an elliptic curve over $\mathbb{Q}$ but is an elliptic curve over $\overline{\mathbb{Q}}$. Although $3x^3 + 4y^3 + 5z^3 = 0$ describes a smooth curve, it does not have any $\mathbb{Q}$-rational points.This definition becomes meaningful if we consider the Tate-Shafarevich group as a set of elliptic curves and a counterexample to the local-global principle. $\endgroup$ Commented Mar 7 at 17:01
  • $\begingroup$ @Piquito Your first statement is false. For example, as commented in my answer, $X^3+Y^3+60Z^3=0$ is rank $0$ elliptic curve over $\Bbb{Q}$ and it is an elliptic curve over $\Bbb{Q}$ because it has a base point $[1:-1:0]$. $\endgroup$ Commented Mar 8 at 16:14
  • $\begingroup$ Okay Pont. I was wrong as you say. I have remembered that the cubics can be either of genus $0$ or genus $1$ and its are elliptic when the genus is $1$ forgetting the essential fact of a rational point playing as zero (can be any other than the mainly used , the point at infinity. I am going to be $86$ years old and this makes me make mistakes frequently. By the way, your $3x^3+4y^3=-5z^3$ is equivalent to $$[9(3x^3)(4y^3)^2+(4y^3-3x^3)^3]^3+[9(3x^3)^2(4y^3)-(4y^3-3x^3)^3]^3=60[-3((3x^3)^2+83x^3)(4y^3)+(4y^3)^2]^3(xyz)^3$$ as you can verify (suite) $\endgroup$
    – Piquito
    Commented Mar 9 at 1:50
  • $\begingroup$ if you don't believe it. This for showing you that I am diminished by my age but not completely lost. $\endgroup$
    – Piquito
    Commented Mar 9 at 1:50
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    $\begingroup$ Hi friend. So cubic's Selmer $3x^3+4y^3+5z^3=0$ becomes elliptic over $\Bbb Q(\sqrt[3]3)$, $\Bbb Q(\sqrt[3]2)$ and $\Bbb Q\left(\sqrt[3]{\frac75}\right)$. Not at all evident the rank equal $0$ because for example for all $A$ integer cube-free, the cubic $x^3+y^3=Az^3$, is elliptic over $\Bbb Q$ and always has an infinity of quadratic points whatever its rank be. Regards. $\endgroup$
    – Piquito
    Commented Mar 9 at 21:24
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COMMENT.-Poincaré has proven in "Sur les proprietés arithmétiques des courbes algébriques. Jour. Math. pures et appliquées 7, pages 161-233 (1901)" that every curve with rational coefficients of degree greater than $3$ and genus $1$ is (birationally) equivalent to a cubic of genus $1$.

In the same way that Hilbert and Hurwitz had proven that any curve defined by a polynomial of degree $n$, with rational coefficients and genus $0$, is equivalent to another curve of degree $n-2$ and genus $0$. Then if the degree is odd the curve is equivalent to a line and have infinitely many rational point and if the curve has even degree it is equivalent to a conic. (Look for this to the book "Diophantine equations" by Mordell (1969).

Hilbert and Hurwitz have closed, with the metioned theorem, the study of genus $0$ and the great importance of elliptic curves, from this perspective, will be that it will close the study of curves of genus $1$. Unfortunately the topic of elliptic curves is far from reach this goal.

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    $\begingroup$ A very celebrated theorem (I have forget the author) ensures that if the genus is greater than $1$ then the curve can have only a finite number of rational points. So it is also closed the study of curves of genus $\ge2$ similarly to what happen with genus $0$ from the perspective of rational points for genus $0$ and greater than $1$. It remains to solve the problem of rational points for genus $1$ which depens of elliptic curves. (Sorry for bad English) $\endgroup$
    – Piquito
    Commented Mar 6 at 16:27
  • $\begingroup$ I remembered few minutes after my answer that "the author" is G. Faltings (Fields Medal) but I did not added what I say now. $\endgroup$
    – Piquito
    Commented Mar 8 at 13:39

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