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?

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    $\begingroup$ Let me see if I get your question right. The correspondence you are looking for is {close points of X} <--> {all morphisms k[X]-->k} the latter have to be $k$-homomorphisms, ie, $k-->K[X]-->k$ the composition from $k$ to $k$ has to be the identity. Now let K be a field extension of k. {Points in X with coordinates in K}<-->{all morphism k[V]-->K} again the latter have to be k-homomorphism. Now we generalize the idea of field extension by {T-points in X} whete $T$ is a S-Scheme. Then we copy-paste of the argument above, and we get a functor $T--Hom(T,X)$. This functor represents X. $\endgroup$ Dec 1, 2009 at 18:18
  • $\begingroup$ Hmm, this seems closer in spirit to what I am after. I will have to mull things over for a bit. You might have just answered my question. $\endgroup$ Dec 1, 2009 at 18:28

3 Answers 3


This is going to be perhaps vague, but I'l try to write down the idea. I've been told that in doing scheme theory, most of the time what is studied is not a scheme but a morphism of schemes $f:X\rightarrow S$. The studied of such a maps can be enriched allowing a chance of "base" $S$. Such a scheme $S$ can be crazy, but I'll impose finite properties to the map $f$ to keep everything under control.

Now, as you say, if we have two $S$-schemes $f:X\rightarrow S$ and $Y\rightarrow S$ then we can consider the fiber product of them (which is a categorical product of $X$ and $Y$ over $S$). Such a process can be though of as replacing the base $S$ by the scheme $Y$. In doing so, the new map $f_Y:X_Y\rightarrow Y$ (standing for $p_2:X\times_{S}Y\rightarrow Y$) may be easier to work with than the map itself $f$. This very construction is generalizing the idea of "extending scalars".

Here is a sort of example of a product described above. Given the point $s\in S$ and its residue field $k(s)$ of the local ring $\mathcal{O}_s$ at that point. Then it is well known that $X_s=X\times_S Spec(k(s))$ has an underlying space the "fiber" $f^{-1}(s)$ (as long as $f$ is "finite"). This space can be considered as an algebraic variety over the field $k(s)$. This way $S$ is the scheme parameterizing the varieties $X_s$ some of which a priori may have different ground fields $k(s)$.

On the other hand, back to the fiber product, in considering a variety $X$ over the field $k$ (scheme of finite type over $k$), we can extend the scalars from $k$ to a field extension $K$. In doing so, we may think of such a variety now over a field extension $K$. Here is where the product comes into play. Vaguely enough, this is saying that if you have an equation and you have solved it over the field $k$ now you might ask yourself if you'd be able to solved it again but over a field extension $K$, here you want to perform an extension of scalars and the ideas written above may help out.

Needless to say that we can consider as well $X\times_k Spec(\overline{k})$ and due to the fact that field $k$ may not have been algebraically closed, such a fiber product can be handful.

Now, to get intuition, let's take a look at an example: $$\pi:Spec \mathbb{Z}[i]\rightarrow Spec(\mathbb{Z})$$

Let $X=Spec\mathbb{Z}[i]$ and notice that the fiber $$X_p=X\times_{\mathbb{Z}}\mathbb{F}_p=Spec(\mathbb{Z}[i]\otimes\mathbb{F}_p)$$

Such a fiber is going to have cardinality 2 if $p\equiv 1 mod(4)$ and cardinality 1 if $p\equiv 3 mod(4)$ (this is the fact that $p=a^2+b^2$ where $a,b\in \mathbb{Z}$ if $p\equiv 1mod 4$). These fibers are $X$ reduce mod $p$. Notice that at the generic point we have $X_0=\mathbb{Q}[i]$.

Now the $T-points$ in $X$ where $T=\mathbb{F}_p[i]$, and all that story give rise to the functor which represents the scheme $X$. If I am not wrong, the info that carries such a functor is nothing but that of the fibers of the map $\pi$. So naively it is like having $X$ fibers wise.


A motivation for the term "over" is that if $k$ is a relatively initial ring, for instance a field, then the algebra and geometry of a scheme "over" $k$ have coefficients in $k$. Long before schemes were defined, you could talk about a curve defined "over" the complex numbers or a ring defined "over" the integers. The realization was that you might as well generalize this concept as much as possible and let $k$ be any scheme.

As you say, it also captures the notion of fibrations or families, and part of the point is that in the end that's not really different. For instance if you have a scheme over the integers $\mathbb{Z}$, is that choosing coefficients or is that a fibration? It's both, because you can specialize to $\mathbb{Z}/p$ for any prime $p$ and that specialization is a fiber.


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.

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    $\begingroup$ "I tried to post this as a comment to Csar's post, but the margin was to small to contain it." Haha:) $\endgroup$ Mar 11, 2010 at 19:07

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