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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.

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up vote 2 down vote accepted

Yes, this is true.

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

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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 9:13
    
@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: terrytao.wordpress.com/2010/01/30/… . 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
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