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:

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)

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,

- 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

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 Feb 22 '15 at 18:58accuratedrawing of the real locus of $y^2=x^5-x$ and see for myself that something (probably associativity) breaks down. $\endgroup$ – Lubin Mar 17 '15 at 16:15