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

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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... –  Ramsey Sep 27 '12 at 17:51
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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? –  IMeasy Sep 27 '12 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. –  Ramsey Sep 27 '12 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? –  IMeasy Sep 27 '12 at 20:11
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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. –  René Sep 27 '12 at 21:14

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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.

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nice example, thank you! –  IMeasy Sep 28 '12 at 8:42
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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. –  Daniel Loughran Sep 28 '12 at 9:31
    
Yes, of course. Thank you, Daniel! –  René Sep 28 '12 at 13:53

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