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I recently came across an interesting phenomenon which confused me slightly, concerning integral points on varieties.

For example, consider $X = \mathbb{A}_{\mathbb{Z}}^{n+1} \setminus \{0\}$, affine $n$-space over $\mathbb{Z}$ with the origin removed. Naively, one would guess that $X(\mathbb{Z})$ is the set of integers $\{ (x_0,x_1,\ldots,x_n) \in \mathbb{Z}^{n+1} \setminus \{0\}\}$.

However, some work that I have been doing on recently with universal torsors has in fact led me to believe that $X(\mathbb{Z})$ should equal $\mathbb{P}^n(\mathbb{Z})$, at least modulo the action of $\mathbb{G}_m$. That is $X(\mathbb{Z})$ is actually the set of integers $\{ (x_0,x_1,\ldots,x_n) \in \mathbb{Z}^{n+1}\setminus \{0\}\}$ such that $\gcd(x_0,x_1,\ldots,x_n)=1$.

Is there a simple explanation for why this is the case?

Thanks! Dan

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This question was given to me as preparation for my Oral Exam! –  H. Hasson Dec 21 '10 at 2:07
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2 Answers

up vote 11 down vote accepted

Your belief is correct. A $\mathbb{Z}$-point has to reduce to an $\mathbb{F}_p$-point for all $p$, which kills examples with gcd > 1.

If you want to make this precise, try writing down an explicit description of X by patching affine pieces. All the essential ideas are already there in $\mathbb{A}_{\mathbb{Z}}^1 \backslash 0$: this is the spectrum of $\mathbb{Z}[X, Y] / (XY - 1)$.

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Thanks for the help David, I understand it now! It needs to be non-zero mod $p$ for all reductions mod $p$. Simple really... –  Daniel Loughran Jun 17 '10 at 12:00
I guess it depends on how on interprets $\mathbf{A}_{\mathbf{Z}}^n$ minus the ``origin," but if one interprets it as the complement of $V((x_1,\ldots,x_n))$, then the $n=1$ case is kind of special, isn't it? For $n\geq 2$, the height of $(x_1,\ldots,x_n)$ is $\geq 2$, so the ring of sections of the complement of $V((x_1,\ldots,x_n))$ is all of $\mathbf{Z}[x_1,\ldots,x_n]$ by "Hartog's lemma," and so the complement cannot be affine. –  Keenan Kidwell Jun 3 '12 at 3:14
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Dear Daniel, here is a detailed explanation, respectfully following the sacred texts (EGA or Hartshorne).

a) First of all, $\mathbb A^{n+1}_{\mathbb Z}$ has no origin, despite our classical intuition! As a substitute, it has the prime (but not maximal) ideal $\mathcal P=(X_1,X_2,...,X_{n+1})$ and corresponding to it the integral subscheme $V=V(\mathcal P)$. And what you want to calculate is the set of $\mathbb Z$-points of $U=\mathbb A^{n+1}\setminus V(\mathcal P)$, i.e. the set of morphisms $Spec(\mathbb Z)\to U$. Let's do that.

b) A morphisms $f: Spec \mathbb Z \to \mathbb A^{n+1}$ corresponds to a morphism of rings $ev_a: \mathbb Z[X_1,X_2,...,X_{n+1}] \to \mathbb Z$ , evaluation of integral polynomials at a tuple $a=(a_1,a_2,...,a_{n+1}) \in {\mathbb Z}^{n+1} $. Call $f=f_a : Spec \mathbb Z \to \mathbb A^{n+1}$ the corresponding morphism.

c) We must ensure that the image of $f$ lies in $U$ i.e. that it is disjoint from $V=V(\mathcal P)$. But the points of $V$ are its generic point $(X_1,X_2,...,X_{n+1})$ and its closed points $\mathcal M_p=(X_1,X_2,...,X_{n+1}, p)$, $p$ a prime. It is enough to show that these closed points are not in the image of $f$. Equivalently, we must show that the fibre of $f$ at $\mathcal M_p$ is empty. Since this fibre is the spectrum of $\mathbb Z / (a_1,a_2,...,a_{n+1},p)$ our condition is that this ring be zero or equivalently that the ideal $(a_1,a_2,...,a_{n+1},p)$ be zero: this will happen exactly if $p$ does not divide all ot the $a_i$'s. Since this must hold for all primes, we get:

Final result The $\mathbb Z$-points of $U=\mathbb A^{n+1}\setminus V(\mathcal P)$ are given by $(n+1)$-tuples of integers whose g.c.d. is 1.

Reminder I have used that the fibre of a morphism of affine schemes $f:SpecB\to Spec A$ at $\mathcal M \in Specmax A$ is $Spec ( B/\mathcal M B) $.

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Thanks Georges! Its nice to see everything worked out explicitly. I tried to do a similar computation and got stuck and hence why I came here... –  Daniel Loughran Jun 17 '10 at 15:03
I think being just a little bit less explicit is better here: A morphism $\mathrm{Spec}S\to\mathrm{Spec}R$ has image in the complement of a closed subscheme $V(I)$ precisely when $SI=S$. In the case at hand $I=(X_1,\dots,X_{n+1})$ so that the map lies in the complement precisely when $(a_1,\dots,a_{n+1})=\mathbb Z$. –  Torsten Ekedahl Jun 17 '10 at 16:14
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