Integer points (very naive question) 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?
 A: Dear unknown, it makes no sense, given a scheme over $\mathbb C$ to ask whether the coordinates of a point are integers . First and foremost because the points of such a scheme have no coordinates ! You might try to embed your scheme in some affine space (say),  but even if you succeed, the coordinates of the corresponding point  depend on the embedding (as your example shows) and so does their arithmetic nature.
This is not so strange after all: the circles in the real plane with equations $x^2+y^2=1$ and $x^2+y^2=3 $ are diffeomorphic but the first has points with rational coordinates and the second hasn't. Differential geometry doesn't see rational points. Neither does algebraic geometry over $\mathbb C$.
EDIT Considering the comments of the OP, I think I should be more explicit in my answer.
My point was the following. Suppose I consider the affine $\mathbb Q$-scheme $X=Spec(A)$ corresponding to the finitely generated $\mathbb Q$- algebra $A$. Then among the points of $X$ there is a perfectly well defined subset $X_0 \subset X$ consisting of $\mathbb Q$- points.Now there are many, many ways of embedding $X$ in $\mathbb A _{\mathbb Q}^n$, corresponding to choices of generators for $A$ .
But whichever embedding you choose, the points of the embedded variety which have rational coordinates are exactly the images of $X_0$ under the embedding. So there is an intrinsic meaning in this case to the phrase "rational points" of $X$. Not so if $X$ is only defined over $\mathbb C$. 
The case of $\mathbb Z$, which is what you actually asked, is similar but as Martin Bright points out in his very clear and informative answer, the naïve analogon is not satisfactory: you have to choose a model and the answer to the question " what are the integer points of $X$" is no longer intrinsic but depends on the model chosen. 
A: 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 $\mathbb{C}$.
A: 
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.
