Motivating Example for Algebraic Geometry/Scheme Theory

I am in the process of trying to learn algebraic geometry via schemes and am wondering if there are simple motivating examples of why you would want to consider these structures.

I think my biggest issue is the following: I understand (and really like) the idea of passing from a space to functions on a space. In passing from $k^n$ to $R:=k[x_1,\ldots,x_n]$, we may recover the points by looking at the maximal ideas of $R$. But why consider $\operatorname{Spec} R$ instead of $\operatorname{MaxSpec} R$? Why is it helpful to have non-closed points that don't have an analog to points in $k^n$? On a wikipedia article, it mentioned that the Italian school used a (vague) notion of a generic point to prove things. Is there a (relatively) simple example where we can see the utility of non-closed points?

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Here's a simple intersection theoretic example.

Take the intersection of the line $y=0$ and the parabola $y=x^2$. Classically, the intersection is a point. But note that there is more to the intersection than just the point; there is the fact that the two curves are tangent at that point. Scheme-theoretically, the intersection is $\operatorname{Spec} k[x,y]/(y,y-x^2) \cong \operatorname{Spec} k[x]/(x^2)$, which has exactly one closed point and exactly one non-closed point. This reflects the tangency. If the intersection were transverse, then the scheme-theoretic intersection would have been just $\operatorname{Spec} k$, which has a single closed point and no non-closed points.

Higher order tangencies can be seen in the scheme-theoretic intersection as well; for example, repeat this exercise with $y=x^3$ in place of $y=x^2$.

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$\operatorname{Spec} k[x]/(x^2)$ has only one point. But this point is "thicker" than $\operatorname{Spec} k$ as it remembers a germ of the tangent line. This is a good example, but for a different thing. It illustrates the usefulness of nilpotents rather than generic points. –  KotelKanim Nov 12 '13 at 15:35

See my answer here for a brief discussion of how points that are closed in one optic (rational solutions to a Diophantine equation, which are closed points on the variety over $\mathbb Q$ attached to the Diophantine equation) become non-closed in another optic (when we clear denominators and think of the Diophantine equation as defining a scheme over $\mathbb Z$).

In terms of rings (and connecting to Qiaochu's answer), under the natural map $\mathbb Z[x_1,...,x_n] \to \mathbb Q[x_1,...,x_n]$, the preimage of maximal ideals are prime, but not maximal.

These examples may give impression that non-closed points are most important in arithmetic situations, but actually that is not the case. The ring $\mathbb C[t]$ behaves much like $\mathbb Z$, and so one can have the same discussion with $\mathbb Z$ and $\mathbb Q$ replaced by $\mathbb C[t]$ and $\mathbb C(t)$. Why would one do this?

Well, suppose you have an equation (like $y^2 = x^3 + t$) which you want to study, where you think of $t$ as a parameter. To study the generic behaviour of this equation, you can think of it as a variety over $\mathbb C(t)$. But suppose you want to study the geometry for one particular value of $t_0$ of $t$. Then you need to pass from $\mathbb C(t)$ to $\mathbb C[t]$, so that you can apply the homomorphism $\mathbb C[t] \to \mathbb C$ given by $t \mapsto t_0$ (specialization at $t_0$). This is completely analogous to the situation considered in my linked answer, of taking integral solutions to a a Diophantine equation and then reducing them mod $p$.

What is the upshot? Basically, any serious study of varieties in families (whether arithmetic families, i.e. schemes over $\mathbb Z$, or geometric families, i.e. parameterized families of varieties) requires scheme-theoretic techniques and the consideration of non-closed points.

(Of course, serious such studies were made by the Italian geometers, by Lefschetz, by Igusa, by Shimura, and by many others before Grothendieck's invention of schemes, but the whole point of schemes is to clarify what came before and to give a precise and workable theory that encompasses all of the contexts considered in the "old days", and is also more systematic and more powerful than the older techniques.)

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Great answer Matt! –  Joachim Mar 27 '13 at 13:17

I think I can at least convince you that it's a good idea to work with non-reduced rings, which aren't really captured by their maximal ideals. The idea is that you get more functors. Just as $\text{Hom}(-, \mathbb{C})$ is the functor which sends a finitely-generated domain over $\mathbb{C}$ to its set of $\mathbb{C}$-points, it turns out that $\text{Hom}(-, \mathbb{C}[x]/x^2)$ sends a finitely-generated domain over $\mathbb{C}$ to its set of $\mathbb{C}$-points together with a choice of tangent vector. In fact this is one way to define the Zariski tangent space. So, on the scheme side, it's a good idea to study morphisms from $\text{Spec } \mathbb{C}[x]/x^2$ to your scheme, and you can't do this if you identify $\mathbb{C}[x]/x^2$ with its set of $\mathbb{C}$-points. (You can't even do this if you identify $\mathbb{C}[x]/x^2$ with its prime ideals; you really need the entire structure sheaf.)

(A really nice application: an algebraic definition of the Lie algebra of an algebraic group.)

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One reason is that you lose the functoriality of $Spec$ if you stick to $MaxSpec$: the inverse image of a maximal ideal is not necessarily maximal. Nevertheless, if you stick to schemes of finite type over a field, this is true (it's basically a version of the Nullstellensatz). In particular, in Serre's FAC paper he defines a "variety" by gluing together regular affine algebraic sets in the sense of classical algebraic geometry. But this is less general. One natural example of a scheme which is not of finite type over a field is simply $Spec \mathbb{Z}$. Then given a scheme $X$ over this (well, admittedly every scheme $X$ is a scheme over $Spec \mathbb{Z}$ in a canonical way), the fibers at the non-closed point of $Spec \mathbb{Z}$ is still interesting and basically amounts to studying polynomial equations over $\mathbb{Q}$ (when $X \to \mathbb{Z}$ is of finite type).
isn't there only one non-closed point of $\text{Spec}(\mathbb{Z})$, namely $(0)$, and the fiber of $X\rightarrow\text{Spec}(\mathbb{Z})$ at $(0)$ amounts to studying things over $\mathbb{Q}$? –  Zev Chonoles Apr 21 '11 at 10:35