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I saw in a document that $y^2 = 1+x^4$ is (the affine equation of) an elliptic curve. Why is it the case? Typically, SAGE tells me it is isomorphic to $y^2 = x^3 - 4x$, which is an elliptic curve with Weierstrass equation, but I don't know how to prove this.

Thank you!

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  • $\begingroup$ Apparently, $y^2=1-x^4$ is also an elliptic curve… Do all equations like $y^2 = $ quartic, give elliptic curves? $\endgroup$ – Alphonse Sep 1 '18 at 10:09
  • $\begingroup$ I found this: math.stackexchange.com/questions/1591990, but we have to be careful about isomorphic/birationally equivalent/isogenous... $\endgroup$ – Alphonse Sep 1 '18 at 10:11
  • $\begingroup$ What definition are you using for isomorphism of curves? $\endgroup$ – Somos Sep 1 '18 at 10:55
  • $\begingroup$ Another example is the elliptic curve $u^3+v^3=w^3$ being $y^2=x^3-432$ via $x=12w/(u+v), y = 36(u-v)/(u+v)$, given p. 53 in Alf van der Poorten's book on Fermat Last theorem. $\endgroup$ – Alphonse Sep 17 '18 at 15:16
  • $\begingroup$ In general, I think that if $y^2 = f(x)$ is an affine model of a smooth projective curve $C$, then the genus of the hyperelliptic curve $C$ is $g$, where the degree of $f$ is $2g+1$ or $2g+2$ (see 1st § of en.wikipedia.org/wiki/Hyperelliptic_curve) $\endgroup$ – Alphonse Oct 21 '18 at 14:43
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A very concrete answer to your question can be found in Exercise 1.15 on page 31 of Silverman and Tate's Rational Points on Elliptic Curves (2nd Edition).

You ask in a comment to your question:

Do all equations of the form $y^2 = \text{quartic}$ give elliptic curves?

The answer is clearly no, since for example the curve $y^2 = x^4$ has a singularity at the origin. Nevertheless, if you add the assumption that the quartic has no repeated roots then the answer is yes, as has already been mentioned in another answer.

In particular, the content of the exercise mentioned above says that if $g(t) \in \mathbb{C}[t]$ is a quartic polynomial, if $\alpha \in \mathbb{C}$ is a root of $g(t)$, and if $\beta \neq 0$ is any number, then the equations \begin{align*} x = \frac{\beta}{t - \alpha} \quad \text{and} \quad y = x^2 u = \frac{\beta^2 u}{(t - \alpha)^2} \end{align*}

give a birational transformation $\phi: \mathcal{Q} \dashrightarrow \mathcal{E}$ between the curve $\mathcal{Q}: u^2 = g(t)$ and the curve $\mathcal{E}: y^2 = f(x)$, where \begin{align*} \phi: \mathcal{Q} &\dashrightarrow \mathcal{E}\\ (t, u) &\mapsto (x, y) = \left(\frac{\beta}{t - \alpha}, \frac{\beta^2 u}{(t - \alpha)^2} \right) \end{align*}

and $$f(x) = g'(\alpha) \beta x^3 + \dfrac{g''(\alpha)}{2!} \beta^2 x^2 + \dfrac{g'''(\alpha)}{3!} \beta^3 x + \dfrac{g^{''''}(\alpha)}{4!} \beta^4$$

is cubic. Moreover, the exercise asks to show that if all the complex roots of $g(t)$ are different, then also the roots of $f(x)$ are distinct, and hence that $\mathcal{Q}: u^2 = g(t)$ is an elliptic curve.


An Example

For instance we can apply this to the curve $u^2 = 1 - t^4$ (the roots of $1 - t^4$ are slightly easier to work with than the roots of $1 + t^4$). In this case $g(t)= 1 - t^4$ has as roots the fourth roots of unity $\pm 1, \pm i$. If we choose $\alpha = 1$ and $\beta = -\dfrac{1}{4}$, then the transformation \begin{align*} x = -\frac{1}{4} \frac{1}{t - 1} \quad \text{and} \quad y = \frac{1}{16} \frac{u}{(t - 1)^2} \end{align*}

gives a birational transformation with the curve $$ y^2 = f(x) = x^3 - \frac{3}{8}x^2 + \frac{1}{16} x - \frac{1}{256}, $$

which is already in Weierstrass form. Moreover, if you want you can depress the cubic by making the change $x \mapsto X - \dfrac{1}{3}\left( -\dfrac{3}{8} \right) = X + \dfrac{1}{8}$ and $y \mapsto Y$, which gives you the equation $$ Y^2 = X^3 + \frac{1}{64} X. $$

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  • $\begingroup$ Just to be sure: $(uv)^2 = v^4 - t^4$ defines a singular quartic in $\Bbb P^2$, which is birationally equivalent to $y^2z = x^3 + xz^2/64$ over $\Bbb Q$. Is it correct? However, the latter is a smooth cubic curve and can't be isomorphic to the former, right? Thank you! $\endgroup$ – Alphonse Sep 25 '18 at 15:08
  • $\begingroup$ Should I ask a new question to get an answer to my above comment? I think it is strongly related to your nice post, so that you may be able to answer it ;-). Many thanks ! $\endgroup$ – Alphonse Sep 26 '18 at 7:34
  • $\begingroup$ (When I wrote "quartic" above, it means that the degree of the projective variety is $4$, as in p. 52 Hartshorne) $\endgroup$ – Alphonse Oct 21 '18 at 14:23
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To me, an elliptic curve over a field $k$ is a (projective, non-singular) genus one curve defined over $k$ with a specified point $O$, also defined over $k$, to serve as an identity in its group. The Weierstrass curve $y^2=x^3+ax+b$ is, if non-singular, an (affine model of an) elliptic curve with the point at infinity the point $O$.

In general, a curve $y^2=f(x)$ with $f$ a quartic having no repeated zeros is an affine model of a non-singular genus one curve. (It will have two points "at infinity"). Is it an elliptic curve? I'd say no, until one chooses a $O$ point. To do that over the field $k$, either one needs to pick a point $(x_0,y_0)$ on the curve with $x_0$, $y_0\in k$ or one must pick a point at infinity at zero. But the points at infinity are defined over $k$ iff $a$ is a square in $k$ where $f(x)=ax^4+\cdots$.

In your examples, $y^2=x^4+1$ is an elliptic curve, where you can choose a point at infinity or $(0,1)$ as $O$. Also $y^2=1-x^4$ is, again choosing $O=(0,1)$ but over $\Bbb Q$, the points at infinity are not defined over $\Bbb Q$.

[I'm assuming $k$ is not of characteristic $2$ throughout.]

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  • $\begingroup$ Thank you. Can you explain (or give a precise reference) why a curve $y^2=f(x)$ with $f$ a quartic having no repeated zeros, gives has genus one ? $\endgroup$ – Alphonse Sep 1 '18 at 11:23
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    $\begingroup$ Apply the Riemann-Hurwitz formula for the genus to the map $E\to\Bbb P^1$ defined by $(x,y)\mapsto x$. @Alphonse $\endgroup$ – Lord Shark the Unknown Sep 1 '18 at 11:32
  • $\begingroup$ Shouldn't we get that the genus is $g=3$, from the genus-degree formula, the degree being $d=4$? $\endgroup$ – Alphonse Sep 1 '18 at 11:37
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    $\begingroup$ @Alphonse As I said, it's not a non-singular quartic. See en.wikipedia.org/wiki/Hyperelliptic_curve $\endgroup$ – Lord Shark the Unknown Sep 1 '18 at 11:43
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    $\begingroup$ $y^2=x^4+1$ is a double cover of $\Bbb P^1$ ramified at the primitive eight roots of unity. $y^2=x^3-4x$ is a double cover of $\Bbb P^1$ ramified at $0$, $\pm 2$ and $\infty$. They'll be isomorphic over $\Bbb C$ if the cross-ratios of these quartets of points are the same. But, I'd trust SAGE on this one; this is very much the sort of maths it was invented for. $\endgroup$ – Lord Shark the Unknown Sep 1 '18 at 16:07
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A partial answer, at least: $$y^2=(1-x^2)(1-k^2 x^2)$$ is the elliptic curve associated with the Jacobi elliptic function $\operatorname{sn}(\cdot,k)$. For $k=i$, you get $y^2=1-x^4$ (see lemniscatic elliptic function). And $1+x^4$ is the same thing as $1-x^4$, up to rotating the complex $x$ plane 45 degrees.

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