MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
This question was given to me as preparation for my Oral Exam! – H. Hasson Dec 21 '10 at 2:07
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 $\mathbb{Z}[X, Y] / (XY - 1)$.

share|cite|improve this answer
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

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) $.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.