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How easy is it to write down genus one curves over $\mathbf Q$ without a rational point?

Can we write down an infinite family?

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    $\begingroup$ It depends on what you mean by "write down." You can argue directly that there must be infinitely many even with the same Jacobian by seeing that $H^1(Gal(\overline{\mathbf{Q}}/\mathbf{Q}), E)$ has arbitrarily large cyclic subgroups. There is a correspondence between these coho classes and principal homogeneous spaces for $E$ (although different classes can be isomorphic as $\mathbf{Q}$-curves). $\endgroup$ – Matt Nov 21 '13 at 4:00
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It's easy enough to write down curves with no $p$-adic points. For example $$ C: X^3+pY^3=p^2Z^3 $$ has no projective solutions over $\mathbb{Q}_p$. Indeed, if it did, then you could assume $X,Y,Z$ to be $p$-adic integers with at least one of them a unit. But reducing modulo $p$, you see that $p|X$, and then reducing modulo $p^2$ you get $p|Y$ and finally the left hand side is divisible by $p^3$, so $p|Z$, which is a contradiction.

But much more interestingly, Poonen writes down completely explicitly an infinite family of genus 1 curves that violate the Hasse principle, i.e. they have points everywhere locally, but no global point.

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If you're looking for curves with points everywhere locally, you could adapt the methods from this paper of Bremner. He uses the arithmetic of the quadratic field generated by the square root of $-29$ to construct a genus 2 curve with no rational points but points everywhere locally. The existence of local points is proven by combining the Riemann Hypothesis over Finite Fields with direct computations for a small number of primes.

The idea is to consider curves of the form $y^2 = px^4 - q$, where the number field generated by a square root of $-q$ has class number 4. $p$ should be chosen so that it splits as a product of nonprincipal ideals in this number field.

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