Timeline for Is there a substitution that relates every Fermat curve to an elliptic curve?
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15 events
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| Dec 31, 2021 at 20:51 | comment | added | Thomas | On the positive side, there is a morphism from the N=7 Fermat equation to an elliptic curve, namely the quotient by the symmetric group on three elements. This actually provides an extremely nice proof of a symmetric generalization of Fermat for that exponent. | |
| Dec 20, 2018 at 11:17 | vote | accept | YiFan | ||
| Dec 20, 2018 at 11:01 | answer | added | François Brunault | timeline score: 19 | |
| Dec 20, 2018 at 10:08 | comment | added | abx | @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. | |
| Dec 20, 2018 at 9:38 | comment | added | François Brunault | 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. | |
| Dec 20, 2018 at 9:24 | comment | added | François Brunault | 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. | |
| Dec 20, 2018 at 7:25 | comment | added | abx | @Sándor Kovács: Oops, you are absolutely right. I misunderstood the question. | |
| Dec 20, 2018 at 7:12 | comment | added | Sándor Kovács | @abx I think you reversed something in your argument. Here is an analog: A given curve can only dominate a finite number of smooth rational curves (aka a $\mathbb P^1$), so according to your argument there are only a countable set of curves altogether, since every curve dominates a $\mathbb P^1$. $\smile$ | |
| Dec 20, 2018 at 7:10 | review | Close votes | |||
| Dec 20, 2018 at 12:44 | |||||
| Dec 20, 2018 at 5:21 | comment | added | abx | A given curve can only dominate a finite number of elliptic curves. So the elliptic curves dominated by some Fermat curve form a countable set. | |
| Dec 20, 2018 at 3:00 | comment | added | Joe Silverman | 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. | |
| Dec 20, 2018 at 2:54 | comment | added | Joe Silverman | @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? | |
| Dec 20, 2018 at 1:04 | comment | added | Adam P. Goucher | Unless I'm missing something, higher-degree Fermat curves have genus greater than 1, which means that they're not birationally equivalent to elliptic curves. | |
| Dec 20, 2018 at 0:40 | review | First posts | |||
| Dec 20, 2018 at 1:48 | |||||
| Dec 20, 2018 at 0:39 | history | asked | YiFan | CC BY-SA 4.0 |