# Is the Hasse principle a birational invariant?

Is the Hasse principle a birational invariant?

It is probably a very trivial question, but I am a beginner in arithmetics.

• I'm not sure how well-posed this questions is. A particular variety has points or it doesn't, so "the Hasse principle holds" is the sort of thing one says about a collection of varieties characterized in some way. For a particular such collection, it makes sense to ask if it is closed under birational isomorphism, but I'm not sure what the question in the title could mean precisely as stated... Commented Sep 27, 2012 at 17:51
• I guess that the answer to my question is very likely to be NO. I assume that for the variety (say projective) A the Hasse principle holds. The projective variety B is birationally equivalent to A, does the Hasse principle hold for B? Commented Sep 27, 2012 at 17:57
• I guess my point is that, for a fixed variety $A$, the statement that "the Hasse principle holds for $A$" has little meaning. This variety either has a rational point or it does not. Commented Sep 27, 2012 at 20:06
• OK sorry. I see what you mean, I was being a little sloppy, you are right. Let's put it this way: say I have a class of varieties for which the principle holds: I pick up one V and find another variety W - not belonging to the same class- that is birationally equivalent to V. Do I expect W to belong to a second class of varieties for which the principle holds? Commented Sep 27, 2012 at 20:11
• Saying that for a variety X over a global field k the HP holds does have meaning: saying HP holds for X means that the implication X has local points everywhere'' $\Rightarrow$ `X has a global point'' is true. The only way in which it could fail if X has local points everywhere but does NOT have a global point. In this way, HP can be said to hold or not hold for any set or class of varieties, including singleton sets.
– R.P.
Commented Sep 27, 2012 at 21:14

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.
• I think for the OP's benefit I should also clarify that for Lang-Nishimura you also need properness. Indeed, given a smooth variety $X$ with a single rational point $x$, the variety $Y=X \setminus x$ is birational to $X$ yet has no rational points. Commented Sep 28, 2012 at 9:31