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Let $\mathbb{Q}^{alg}$ be the algebraic closure of the rationals. Given a point $P\in \mathbb{A}^n(\mathbb{Q}^{alg})$, $P = (a_1,\dots,a_n)$, we define the degree of $P$ to be the degree of the minimal field extension of $\mathbb{Q}$ over which $P$ is defined: $\text{deg}(P) = [\mathbb{Q}(a_1,\dots,a_n):\mathbb{Q}]$.

If a variety $X$ in $\mathbb{A}^n(\mathbb{Q}^{alg})$ has infinitely many points, must it have infinitely many points of bounded degree? That is, is there a positive integer $d$ such that $\{P\in X:\text{deg}(P)\leq d\}$ is infinite?

It seems like this should be an easy consequence of some more general theorem, but my knowledge is limited.

Alex Kruckman
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1 Answers1

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Yes, this is true.

Hint: we may assume that $X$ is irreducible (take any one irreducible component of positive dimension). Apply Noether Normalization.

Pete L. Clark
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  • Pete, do you feel like there ought to be a more general model-theoretic argument here? – Qiaochu Yuan Feb 14 '11 at 09:13
  • @Qiaochu: I'm not sure what you have in mind here. Could you say a little more? – Pete L. Clark Feb 14 '11 at 12:56
  • something like the arguments here: http://terrytao.wordpress.com/2010/01/30/the-ultralimit-argument-and-quantitative-algebraic-geometry/ . I haven't really thought about this, though. – Qiaochu Yuan Feb 14 '11 at 14:00
  • @Qiaochu: hmm, okay, I sort of see what you're getting at. But I haven't read that post (as usual, really an expository article unto itself) of Tao's in any detail. I'll have to get back to you on this... – Pete L. Clark Feb 14 '11 at 15:41