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I know that the points of an elliptic curve over $\mathbb{Q}$, $\mathbb{R}$ or other field $K$ form a group, particularly the most common example to explain the naive way is with this curve $y^2=x^3-x$, internally what is happening is that line with coefficients over the ground field intersects with $K$-rational points, my questions are:

  1. Why a curve $y^2=x^5-x$ which is of genus 2 and "looks like" $y^2=x^3-x$ if it is taken over $\mathbb{R}$ what is going to fail in the group operation? (every line must intersect 3 $\mathbb{R}$-rational points (with counted multiplicity) , what I believe is that 'associativity' is going to fail... but I dont see it, but I know the points doesn't form a group (The jacobian does)

  2. Are the points of the last genus 2 curve a set which forms a group under the usual line-tangent rule for some (finite) field?, I think no... but why

I know that the points of this curve does not form a group, but in $\mathbb{R}$ the curve 'looks' the same for having a similar naive definition for the explanation of the group operation,

  1. If I 'force' the group operation in this curve calculating $nP$ (generating it) for some point $P\in H(\mathbb{R})$ where $H(x,y)=y^2-x^5+x$ do I get a group? (Again... It shouldnt but why?)

Thanks

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  • $\begingroup$ In general a curve $C$ of genus $g>0$ has a Jacobian $J(C)$, which is a group of dimension $g$. If $C$ has a rational point then it embeds into $J(C)$, but not as a subgroup except in the $g=1$ case of an elliptic curve in which case $C$ is its own Jacobian. $\endgroup$
    – Noam D. Elkies
    Commented Feb 22, 2015 at 18:58
  • $\begingroup$ And, aren't all genus 2 curves over any field hyperelliptic? $\endgroup$
    – Qfwfq
    Commented Feb 22, 2015 at 19:46
  • $\begingroup$ @Qfwfq: Yes, they are. $\endgroup$
    – Michael Stoll
    Commented Feb 22, 2015 at 19:58
  • $\begingroup$ My inclination would be take an accurate drawing of the real locus of $y^2=x^5-x$ and see for myself that something (probably associativity) breaks down. $\endgroup$
    – Lubin
    Commented Mar 17, 2015 at 16:15
  • $\begingroup$ Lubin, yes, that's what I'm in fact asking, As @Michael Stoll replied, there are more intersection points, and there is not a way to build a group with the points, but what I ask is that if I take two P,Q $\mathbb{R}$-rational points on $y^2=x^5-x$, the line between them is going to intersect another $\mathbb{R}$-rational point, (can this be assumed? I think thats the main problem) If I take that point naively to be -(P+Q), what I get ?, does associativity fails? What if I fix a $\mathbb{R}$-rational point R not lying in the x axis and I generate $nR$ naively also nR, that should fail also. $\endgroup$
    – Eduardo R. Duarte
    Commented Oct 9, 2015 at 14:04

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Basically, the problem is that there are really five intersection points with your line (two of them are complex non-real). If you could make a group in the same way as for elliptic curves, then this would work over arbitrary fields, including the complex numbers. But when you have three more points of intersection instead of just one, you don't know which one to take.

Also, it is a result of topology that a topological group has to have vanishing Euler characteristic, but the Riemann surface you get here from the complex points has Euler characteristic $-2$, so there is not even a continuous (in the complex topology) group law on it. (Of course, you can take a homeomorphism of the real points with $S^1 \times {\mathbb Z}/2{\mathbb Z}$ and so force a continuous group law on the real points, but this is not what you are asking.)

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