TITLE: Intuition about schemes over a fixed scheme
QUESTION [15 upvotes]: I am taking a first course on Algebraic Geometry, and I am a little confused at the intuition behind looking at schemes over a fixed scheme.  Categorically, I have all the motivation in the world for looking at comma categories, but I would like to make sense of this geometrically.
Here is one piece of geometric motivation I do have: A family of deformations of schemes could be thought of as a morphism $X \rightarrow Y$, where the fibers of the morphism are the schemes which are being deformed, and these are indexed by the scheme Y.
This is all well and good, but I am really interested in Schemes over $Spec(k)$ are thought of as doing geometry "over" $k$.  I know that their is a nice "schemeification functor" taking varieties over $k$ to schemes over $k$, but this is somewhat unsatisfying. All that I see algebraically is that $k$ injects into the ring of global sections of the structure sheaf of $X$, but this does not seem like much of a geometric condition...
Any help would be appreciated. 
EDIT:  The answers I have received so far have been good, but I am looking for something more.  I will try to flesh out an example that give the same style of answer that I am asking for:
The notion of a k-valued point:  Every point $(a,b)$ of the real variety $x^2 + y^2 - 1 = 0$ has a corresponding evaluation homomorphism $\mathbb{R}[x,y] \rightarrow \mathbb{R}$ given by $x \mapsto a$ and $y \mapsto b$.  Since $a^2 + b^2 - 1 = 0$, this homomorphis factors uniquely through $\mathbb{R}[x,y]/(x^2 + y^2 -1)$. So the real valued points of the unit circle are in one to one correspondence with the homomorphisms from $\mathbb{R}[x,y]/(x^2 + y^2 -1)$ to $\mathbb{R}$.  Similarly, the complex valued points are in one to one correspondence with the homomorphisms from $\mathbb{R}[x,y]/(x^2 + y^2 -1)$ to $\mathbb{C}$.  Actually, points of the unit circle valued in any field $k$ are going to be in one to one correspondence with homomorphisms from $\mathbb{Z}[x,y]/(x^2 +y^2 -1)$ to $k$.
Dualizing everything in sight, we are saying that the $k$- valued points of the scheme $Spec(\mathbb{Z}[x,y]/(x^2 +y^2 -1))$ are just given by homomorphisms from the one point scheme $Spec(k)$ into $Spec(\mathbb{Z}[x,y]/(x^2 +y^2 -1))$.  
EDIT 2: Csar's comment comes very close to answering the question for me, and I will try and spell out the ideas in that comment a little better. I wish Csar had left his comment as an answer so I could select it.
So it seems to me that the most basic reason to think about schemes over a field $k$ is this: 
I already spelled out above why $k$-valued points of a scheme are important.  But a lot of the time, a morphism from $Spec(k)$ to $X$ will point to a generic point of $X$, not a closed point.  Different morphisms can all point to the same generic point.  For instance the dual of the any injection $\mathbb{Z}[x,y] \rightarrow \mathbb{C}$ (of which there are many), will all "point" to the generic point of $\mathbb{Z}[x,y]$.  
On the other hand if we are looking at $\mathbb{Q}[x,y]$, this is a $\mathbb{Q}$ scheme.  $Spec(\mathbb{Q})$ is also a $\mathbb{Q}$ scheme via the identity map.  A morphism of $\mathbb{Q}$ schemes from  $Spec(\mathbb{Q})$ to $\mathbb{Q}[x,y]$ will correspond to a closed $\mathbb{Q}$ -valued point!  So this is a nice geometric reason for looking at schemes over a fixed field $k$: $k$ morphisms of $Spec(k)$ to $X$ correspond to $k$-valued closed points of $X$.
It will take some work for me to internalize the vast generalization that S-schemes entail, but I think this is a good start.  Does everything I said above make sense?

REPLY [5 votes]: I tried to post this as a comment to Csar's post, but the margin was to small to contain it.
I understand what you are trying to say here, but this style of answer isn't what I am exactly looking for.  Extension of scalars should not be vague. Here is an explicit example (I am sticking with circles as per my edit.):  Say we are looking at rational points of the circle. Then naturally we are interested in $Spec(\mathbb{Q}[x,y]/(x^2+y^2-1))$.  Say we want to instead look at complex valued points.  Well $Spec(\mathbb{Q}[x,y]/(x^2+y^2-1))$ is  a scheme over $\mathbb{Q}$, and $\mathbb{C}$ is a scheme over $\mathbb{Q}$ (the dual of the inclusion maps give the appropriate maps here).  When you take the fiber product to get a scheme over $Spec(\mathbb{C})$, this corresponds to a pushout in the category of rings, i.e. a tensor product. So the fiber product is $Spec(\mathbb{Q}[x,y]/(x^2+y^2-1) \otimes_\mathbb{Q} \mathbb{C})$ which is isomorphic to $Spec(\mathbb{C}[x,y]/(x^2+y^2-1))$, i.e. exactly what you would expect.
I would like something this explicit explaining the geometric meaning of a scheme over a fixed scheme.