TITLE: Integer points (very naive question)
QUESTION [15 upvotes]: Well, I don't have any notion of arithmetic geometry, but I would like to understand what arithmetic geometers mean when they say "integer point of a variety/scheme $X$" (like e.g. in "integer points of an elliptic curve").
Is an integer point just defined as a morphism from $Spec\mathbb{Z}$ into $X$?
Suppose $X$ is given a structure of variety over, say, $Spec\mathbb{C}$; does the notion of integer point interact with this structure?
How is all of this related to finding the integer solutions of a concrete polynomial equation (perhaps with integer coefficients)?
Now a very, very, naif question. Suppose you have the plane $\mathbb{A}^{2}_{\mathbb{C}}$ and draw two lines on it: $X=\{x=0\}$ and $X'=\{x=\pi\}$ (with the obvious reduced induced closed subscheme structure). Well, $X$ and $X'$ are clearly isomorphic as schemes over $\mathbb{C}$ (hence as schemes). But, if having integer points is somehow related to finding integer solutions to equations, how do you explain that $X$ (as embedded in $\mathbb{A}^{2}$) has plenty of points with integer coordinates, while $X'$ has none? 
This to mean: how can the intrinsic notion of a morphism from $Spec\mathbb{Z}$ be possibly related to finding solutions to concrete (coordinate dependent) equations?

REPLY [22 votes]: To make sense of the notion of integer points, your scheme should be defined over $\mathbb{Z}$. What do we mean by that? Of course we should not ask for a structure map tp $Spec(\mathbb{Z})$, since every scheme has one such map. The right notion is the following.
Let $X$ be a scheme over $\mathbb{C}$; so by definition we have a structure map $X \to \mathop{Spec} \mathbb{C}$. Then we say that $X$ is defined over $\mathbb{Z}$ is there exists a scheme $X_{\mathbb{Z}}$ over $\mathbb{Z}$ such that $X$ is the base change of $X_{\mathbb{Z}}$ to $\mathbb{C}$, i. e. $X \cong X_{\mathbb{Z}} \times_{\mathbb{Z}} \mathop{Spec} \mathbb{C}$.
Now for such a scheme an integral point is a map $\mathop{Spec}\mathbb{Z} \to X_{\mathbb{Z}}$ such that the composition with the structure map is the identity. Note that the same can be done for every ring $A$ in place of $\mathbb{Z}$.
With this definition, the line $\{ x = 0 \}$ is defined over $\mathbb{Z}$, but the line $\{ x = \pi \}$ is not, basically because there is no way to generate its ideal with equations having integer coefficients. So your problem does not arise anymore.
EDIT: Abstractly of course the two lines are isomorphic over $\mathbb{C}$, so the line $r = \{x = \pi \}$ actually has a model over $\mathbb{Z}$. The problem is that this model is not compatible with the inclusion in $\mathbb{A}^2$, that is, there will be no map $r_{\mathbb{Z}} \to \mathbb{A}^2_\mathbb{Z}$ whose base change is the inclusion of $r$ into $\mathbb{A}^2$. In order to have this, you would have to ask that the ideal of $r$ in $\mathbb{A}^2$ should be generated by polynomials with integer coefficients.
As for your second question, there can be different models, that is, nonisomorphic schemes over $\mathbb{Z}$ which become isomorphic after base change to $\mathbb{C}$. So before discussing the existence of integral points, you have to FIX a model, and the points will in general depend on the model.
For instance take the two conics $\{ x^2 + y^2 = 2 \}$ and $\{ x^2 + y^2 = 3 \}$. Both have an obvious choice of a model, given by the inclusion in $\mathbb{A}^2$; moreover they are isomorphic over $\mathbb{C}$. But the integral points on the first one are $(\pm 1, \pm 1)$, while the second has none.
Finally you consider the possibility that the structure over $\mathbb{C}$ is not relevant. This is false: the base change $X_\mathbb{Z} \times_{\mathbb{Z}} \mathop{Spec} \mathbb{C}$ is endowed with a natural map to $\mathop{Spec} \mathbb{C}$, and we ask for the isomorphism with $X$ to be over $\mathbb{C}$.

REPLY [12 votes]: Is an integer point just defined as a morphism from Spec ℤ to X?

Yes, but in order for there to be any, your scheme has to be genuinely "over ℤ", that is, equipped with a surjective map to Spec ℤ.  To see what a difference this makes, consider affine space $\mathbb{A}^1_{\mathbb{Q}} = \textrm{Spec} \mathbb{Q}[t]$.  A morphism from Spec ℤ to $\mathbb{A}^1_\mathbb{Q}$ is a homomorphism $\mathbb{Q}[t] \to \mathbb{Z}$, but there aren't any (the image of such a homomorphism would have to contain a copy of ℚ).
So to talk about "integer points" of $\mathbb{A}^1_\mathbb{Q}$, you need to choose a model over ℤ, that is, a scheme $\mathcal{X}$ over Spec ℤ with an isomorphism $\mathcal{X} \times_{\textrm{Spec} \mathbb{Z}} \textrm{Spec} \mathbb{Q} \cong \mathbb{A}^1_\mathbb{Q}$; then you can talk about integer points of $\mathcal{X}$.  A likely candidate is $\mathbb{A}^1_\mathbb{Z} = \textrm{Spec} \mathbb{Z}[t]$, which does indeed have a ℤ-valued point for each element of ℤ, as you might expect.  Now you can call a point of $\mathbb{A}^1_\mathbb{Q}$ an integer point if it extends to a ℤ-valued point of $\mathcal{X}$.  The notion of "integer point" you get totally depends on the model you choose; it's not intrinsic to the original scheme over ℚ.
What makes this a bit more confusing is that we're used to dealing with projective varieties; and, for proper schemes $\mathcal{X}$ over ℤ, every ℚ-valued point extends to a ℤ-valued point; this is an easy exercise using the valuative criterion of properness.  For projective varieties this is fairly obvious: for example, any ℚ-valued point of $\mathbb{P}^1_\mathbb{Q}$ can be represented as $(x:y)$ with x, y integers, so extends to a ℤ-valued point of $\mathbb{P}^1_\mathbb{Z}$.
To see how these two notions fit together, imagine $\mathbb{A}^1_\mathbb{Z}$ embedded in $\mathbb{P}^1_\mathbb{Z}$ as usual; the complement is a divisor D corresponding to the ℤ-point $(1:0)$.  (I'm assuming that you are familiar with what schemes over ℤ look like – spend a while studying the nice pictures in Mumford's Red Book if not.)  A point $(x:y) \in \mathbb{P}^1_\mathbb{Q}(\mathbb{Q})$ lies in $\mathbb{A}^1_\mathbb{Q}(\mathbb{Q})$ if and only if $y \neq 0$.  Choosing $x$ and $y$ to be integers, we extend this point to a point $(x:y) \in \mathbb{P}^1_\mathbb{Z}(\mathbb{Z})$.  Now $(x:y)$ comes from an integer point of $\mathbb{A}^1_\mathbb{Z}$ if and only if the 1-dimensional subscheme of $\mathbb{P}^1_\mathbb{Z}$ corresponding to the point $(x:y)$ doesn't meet the divisor D.  For any prime p, if $p \mid y$ then $(x:y)$ intersects D in the point $(1:0) \in \mathbb{P}^1_{\mathbb{F}_p}$.  So $(x:y)$ lies in $\mathbb{A}^1_\mathbb{Z}$ if any only if no prime divides $y$, and so $y = \pm 1$, which is exactly as you would hope.