Is the Hasse principle a birational invariant?
It is probably a very trivial question, but I am a beginner in arithmetics.
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Sign up to join this communityIs the Hasse principle a birational invariant?
It is probably a very trivial question, but I am a beginner in arithmetics.
In this generality, the answer is no. The projective curve $X$ given by $2y^2z^2 = x^4 - 17z^4$ over the rationals satisfies the HP, since it has local points everywhere (the affine part $z \neq 0$ is given by $2y'^2=x'^4-17$, which is the famous Reichardt-Lind equation which is known to be everywhere locally, but not globally, soluble) and it has the unique rational point $(0:1:0)$. However, this point is singular: so now consider the normalization $X'$ of $X$: it has two points above $(0:1:0)$, neither of which is rational. By the parenthetical remark, $X'$ has local points everywhere, but it doesn't have rational points: therefore $X'$ does not satisfy the HP. Also, $X'$ is birational to $X$, being its normalization.
If you restrict to smooth varieties however, the answer is yes: by Lang-Nishimura, if $X$ and $X'$ are smooth varieties over any field $k$ that are birational to each other, then $X$ has a $k$ point iff $X'$ does.