# Is there a substitution that relates every Fermat curve to an elliptic curve?

I asked this question on MSE but didn't get any response, so I'm asking here. I apologize in advance if this question is not research level.

A Fermat Curve of degree $$n$$ is the set of solutions to $$x^n+y^n=z^n$$, $$x,y,z\in \mathbb R$$. In this question, the OP provides a substitution which relates a Fermat Curve of degree $$n=3,4$$ to two different elliptic curves. To transform the Fermat Curve of degree $$3$$, the substitutions $$a=\frac{12z}{x+y},\quad b=\frac{36(x-y)}{x+y}$$ produce $$b^2=a^3-432$$, an elliptic curve. Similarly for the Fermat Curve of degree $$4$$, the substitutions $$a=\frac{2(y^2+z^2)}{x^2},\quad b=\frac{4y(y^2+z^2)}{x^3}$$ give $$b^2=a^3-4a$$. However, the substitutions used are not at all obvious, which leads me to wonder,

Is there a similar substitution which can relate a Fermat curve of arbitrary degree to an elliptic curve?

This is equivalent to asking whether there is always a nonconstant morphism from a Fermat curve to an elliptic curve.

• @AdamP.Goucher The OP is asking for a morphism from the Fermat curve onto an elliptic curve, not an isomorphism. Equivalently, does the Jacobian of the Fermat curve have an elliptic factor? – Joe Silverman Dec 20 '18 at 2:54
• The cubic Fermat curve is a smooth cubic in $\mathbb P^2$, so has genus 1. It has the rational point $[1,-1,0]$. Hence it is isomorphic to an elliptic curve given by a Weierstrass equation. Finding the transformation is standard. The quartic Fermat curve maps 2-to-1 to the curve $C:u^4+1=v^2$ via $u=x/y$ and $v=z^2/y$. The curve $C$ also has genus 1 and a rational point $(u,v)=(0,1)$, hence it too can be mapped to an elliptic curve in Weierstrass form (using 19th century formulas!) . For higher $n$, there are lower genus curves that the Fermat curve maps to, but they're generally not elliptic. – Joe Silverman Dec 20 '18 at 3:00
• The Jacobian of the Fermat curve decomposes (up to isogeny) as a product of CM abelian varieties. Accordingly the motive of the Fermat curve $X_N$ decomposes as a direct sum of motives associated to Hecke characters of the cyclotomic field $\mathbf{Q}(\zeta_N)$ (see Otsubo's work e.g. On special values of Jacobi-sum Hecke $L$-functions). So your question is about rationality properties of these Hecke characters. – François Brunault Dec 20 '18 at 9:24
• I don't know the answer to the question but here is a reference: Koblitz, Rohrlich, Simple factors in the Jacobian of a Fermat curve. If I understand Theorem 2 correctly, then if $N \geq 5$ is a prime $\equiv 2 \textrm{ mod } 3$ then the simple factors of the Jacobian of $X_N$ all have dimension $(N-1)/2$. So for exemple $X_5$ and $X_{11}$ do not map to any elliptic curve. – François Brunault Dec 20 '18 at 9:38
• @FrançoisBrunault: I think you should write your comment as an answer, this is actually quite nontrivial. The paper you quote is freely available at https://cms.math.ca/openaccess/cjm/v30/cjm1978v30.1183-1205.pdf. – abx Dec 20 '18 at 10:08

As a particular case of Theorem 2 there, if $$N \geq 5$$ is a prime $$\equiv 2 \textrm{ mod } 3$$ then the simple factors of the Jacobian $$J_N$$ of the Fermat curve $$X_N$$ all have dimension $$(N−1)/2$$. So for example, the Fermat curves $$X_5$$ and $$X_{11}$$ do not map to any elliptic curve.
The motive of the Fermat curve $$X_N$$ has been extensively studied: it decomposes as a direct sum of motives associated to Hecke characters of the cyclotomic field $$\mathbf{Q}(\zeta_N)$$, see e.g. Otsubo's work On special values of Jacobi-sum Hecke $$L$$-functions. Note also that every elliptic factor of a Fermat curve must have complex multiplication, essentially because $$X_N$$, and thus its Jacobian $$J_N$$, admits an action of the rather large group $$\mu_N \times \mu_N$$, where $$\mu_N$$ is the group of $$N$$-th roots of unity in $$\mathbf{C}$$. More generally, the factors of $$J_N$$ are CM abelian varieties.