One of the standard facts in algebraic geometry is that a projective scheme is proper. In the language of varieties, one says that the image of a projective variety is closed. The precise statement one proves is that:

Theorem 1 Let ${Y}$ be any variety over the algebraically closed field ${k}$. Let ${Z \subset \mathbb{P}^n_k \times Y}$ be a closed subset. Then the projection of ${Z}$ to ${Y}$ is closed.

This statement is sometimes phrased as saying that ${\mathbb{P}_k^n}$ is “complete.” In many ways, it is a compactness statement. Recall that a compact space ${X}$ has the property that the map ${X \times Y \rightarrow Y}$ for any ${Y}$ is a closed map. The converse is also true if there are reasonable assumptions (I think locally compact Hausdorff on ${X}$ will do it). Of course, these reasonable assumptions don’t apply to a variety with the Zariski topology, but they do if we are working with a variety (say, a quasiprojective variety) over ${\mathbb{C}}$ and we can define the complex topology. And in fact, it turns out that projective varieties are indeed compact in the complex topology. The scheme-theoretic version is a little different. First, the theorem for varieties said that the object ${\mathbb{P}_k^n}$ was “complete” in a certain sense. But the philosophy of Grothendieck is to consider not so much schemes but morphisms of schemes, and to do everything in a relative context. The idea is:

Definition 2 A morphism ${f: X \rightarrow Y}$ is proper if it is separated, of finite type, and universally closed (i.e. any base change is a closed morphism).

I don’t really want to define separated in this post. On the other hand, I should explain what’s going on for quasiprojective varieties over an algebraically closed fields. In this case, the conditions of finite type and separated are redundant. The key condition is universal closedness, which won’t be satisfied for a general morphism. For instance, to say that ${\mathbb{P}^n_k \rightarrow \mathrm{Spec} k}$ is universally closed implies the first lemma about closedness of the projection.

Anyway, the scheme-theoretic version of the completeness of projective space is:

Theorem 3 The map ${\mathbb{P}^n_{\mathbb{Z}} \rightarrow \mathrm{Spec} \mathbb{Z}}$ is proper for any ${n \geq 0}$.

Here ${\mathbb{P}^n_{\mathbb{Z}}}$ is defined in a purely algebraic way, as the “${\mathrm{Proj}}$” of the graded algebra ${\mathbb{Z}[x_0, \dots, x_n]}$. I will assume that readers either know what this means or will look it up in Hartshorne or elsewhere. It is fairly easy to show directly that ${\mathbb{P}^n_{\mathbb{Z}}}$ is separated (i.e., separated over ${\mathrm{Spec} \mathbb{Z}}$). In fact, more generally, the ${\mathrm{Proj}}$ of any ring is a separated scheme. The reason is that a scheme ${X}$ is separated if and only if there is an open affine cover ${\left\{U_i\right\}}$ of ${X}$ such that each intersection ${U_i \cap U_j}$ is affine and ${\Gamma(\mathcal{O}_X, U_i), \Gamma(\mathcal{O}_X, U_j)}$ generate ${\Gamma(\mathcal{O}_X, U_i \cap U_j)}$ as a ring; this is what you get when you translate the separatedness condition

$\displaystyle X \rightarrow X \times_{\mathbb{Z}} X \quad \mathrm{is \ a \ closed \ immersion}.$

For the ${\mathrm{Proj}}$ of a graded ring, we can just use basic open affines as the ${U_i}$. The hard part is the universal closedness. One way is to use the highly useful valuative criterion, as in Hartshorne. This has the benefit of also giving separatedness—though it doesn’t tell you that the ${\mathrm{Proj}}$ of any graded ring is separated. Nonetheless, there is an elementary way to prove this directly as well. I want to explain that.

The statement for varieties

To approach this, let us start by reviewing the proof for varieties. In particular, we shall prove lemma 1, which goes under the name “elimination theory.” The first standard reduction is to observe that being closed is a local property. We can cover ${Y}$ be an open affine cover, so we can just assume that ${Y }$ is affine. Also, since ${Y}$ is imbedded in ${\mathbb{A}^m}$, it suffices to show that the projection

$\displaystyle \mathbb{P}^n \times \mathbb{A}^m \rightarrow \mathbb{A}^m$

is a closed map. So let ${Z \subset \mathbb{P}^n \times \mathbb{A}^m}$ be a closed subset. This means that the set is defined by some equations

$\displaystyle F_\alpha( \mathbf{x}, \mathbf{y})=0,$

where ${\mathbf{x} = (x_0, \dots,x_n), \mathbf{y} = (y_1, \dots, y_m)}$. Here each ${F_\alpha}$ is homogeneous in ${\mathbf{x}}$. To say that the image is closed is the same as saying as the set of ${y \in \mathbb{A}^m}$ such that

$\displaystyle F_\alpha(\cdot, y)$

has no nontrivial root is open in ${\mathbb{A}^m}$. But by the homogeneous Nullstellensatz, this implies that the ${F_\alpha(\mathbf{x}, y)}$ generate a power of the irrelevant ideal in ${k[\mathbf{x}]}$. In particular, consider the ideal ${I = (F_{\alpha}(\mathbf{x}, \mathbf{y})) \subset k[\mathbf{x}, \mathbf{y}]}$. Then consider the base-change

$\displaystyle k[\mathbf{x}, \mathbf{y}] \rightarrow k[\mathbf{x}], \quad \mathbf{y} \rightarrow y.$

For simplicity, we shall denote ${R = k[\mathbf{x}, \mathbf{y}]}$, and ${\mathfrak{m}_y}$ to be the ideal in ${k[\mathbf{y}]}$ vanishing at ${y}$. We have then that ${I \otimes R/(\mathfrak{m}_y R) \supset (R_+)^N \otimes R/\mathfrak{m}_y}$ for some ${N}$. This is precisely the statement that the ${F_\alpha(\mathbf{x}, y)}$ generate a power of the irrelevant ideal. In particular, if ${N}$ denotes the set of homogeneous polynomials of ${\mathbf{x}}$-degree ${N}$ in ${R}$, we have

$\displaystyle I \cap R_N + \mathfrak{m}_y R_N = R_N.$

This is the same thing as saying that

$\displaystyle \mathfrak{m}_y (R_N/I \cap R_N) = R_N/I \cap R_N.$

The fact that ${R_N}$ is a finite module over the ring ${k[\mathbf{y}]}$ implies by Nakayama that the localized module ${(R_N/I \cap R_N)_{\mathfrak{m}_y}}$ is zero. In particular, there is ${f \notin \mathfrak{m}_y}$ but in ${k[\mathbf{y}]}$ such that ${f(R_N / I \cap R_N) =0}$. What does this buy us? Well, it means that on the open set ${D(f) \subset \mathbb{A}^m}$ where ${f}$ is invertible, we have that ${I \cap R_N}$ is just ${R_N}$. So ${Z}$ can have no points lying over ${D(f)}$. This is the open neighborhood of ${y}$ not containing anything in the image of ${Z}$ that we wanted.

0.2. The scheme case

OK. Well, it turns out that pretty much that entire proof carries over to the case of schemes. In particular, we can show:

Proposition 4 Suppose ${A}$ is a commutative ring. Then the map$\displaystyle \mathbb{P}^n_A = \mathrm{Proj} A[x_0, \dots, a_n] \rightarrow \mathrm{Spec} A$

is a closed map.

This is enough to imply properness of ${\mathbb{P}^n_{\mathbb{Z}} \rightarrow \mathrm{Spec} \mathbb{Z}}$ (and consequently that of ${\mathbb{P}^n_{A} \rightarrow \mathrm{Spec} A}$ as properness is preserved under base change). The reason is that to show that ${\mathbb{P}^n \times Y \rightarrow Y}$ is closed for any scheme ${Y}$, it reduces by covering ${Y}$ by open affines to the case of ${Y}$ affine itself. To prove this, let’s try to imitate the previous proof. Consider a closed subscheme ${Z \subset \mathbb{P}^n_A}$. One way to get such a subscheme is to consider a homogeneous ideal ${I \subset A[x_0, \dots, x_n]}$ and consider the inclusion

$\displaystyle Z = \mathrm{Proj} A[x_0, \dots, x_n]/I \rightarrow \mathrm{Proj} A[x_0, \dots, x_n].$

Then the closed subset ${Z}$ is just the set of homogeneous prime ideals in ${A[x_0, \dots, x_n]}$ containing ${I}$ (provided you think of ${\mathrm{Proj}}$ via homogeneous primes). So let’s suppose that a point in ${A}$, represented by a prime ${\mathfrak{p} \in \mathrm{Spec} A}$, has trivial fiber over ${Z}$. We want to show that there is an open neighborhood of ${\mathfrak{p}}$ which has trivial fiber. Well, first of all, we can make this more algebraic. To say that ${\mathfrak{p}}$ is not in the image of ${Z}$ is to say that ${Z \times_{A} k(\mathfrak{p}) = \emptyset}$, where ${k(\mathfrak{p} )}$ is the quotient field at ${\mathfrak{p}}$ and we have the standard morphism

$\displaystyle \mathrm{Spec} k(\mathfrak{p}) \rightarrow \mathrm{Spec} A.$

However, the process of forming ${\mathrm{Proj}}$‘s plays nice with base-change. The precise statement is that if ${A \rightarrow B}$ is a morphism of rings, then the map ${S \rightarrow S \otimes_A B}$ of graded rings induces a cartesian diagram between the appropriate Proj’s and Spec’s.

Probably the easiest way to see this is to use the explicit standard open cover of ${\mathrm{Proj} A}$ by open affines. You can look this up in EGA. Anyway, back to the main proof. So we know that the fiber of ${\mathrm{Proj} A}$ over ${\mathfrak{p}}$ is ${\mathrm{Proj} A \otimes k(\mathfrak{p})}$, and similarly the fiber of ${\mathrm{Proj} A/I = Z}$ is ${\mathrm{Proj} A/I \otimes_A k(\mathfrak{p})}$. To say that the proj of something is zero is to say that it vanishes in high degrees. This means that

$\displaystyle I_N \otimes_A k(\mathfrak{p}) \rightarrow A[x_0, \dots, x_n]_N \otimes_{A_0} k(\mathfrak{p})$

is surjective for large ${N}$. (Here the subscript ${N}$ means degree ${N}$.) By Nakayama’s lemma and the finite generation of ${A[x_0, \dots, x_n]_N}$, we find that ${(I_N)_{\mathfrak{p}} = (A[x_0,\dots, x_n]_N)_{\mathfrak{p}}}$. There is thus ${f \notin \mathfrak{p}}$ such that

$\displaystyle (I_N)_f = (A[x_0, \dots, x_n])_f.$

I claim that the image of ${Z}$ doesn’t intersect ${D(f)}$. Indeed, to see this, we consider the fibered product ${Z \times_A D(f)}$; this is ${\mathrm{Proj} A_f[x_0, \dots, x_n]/I_f}$. But the ideal ${I_f \subset\mathrm{Proj} A_f[x_0, \dots, x_n]}$ contains all the homogeneous elements of degree ${N}$. Since it is an ideal, it contains all homogeneous elements of degree ${\geq N}$. So

$\displaystyle A_f[x_0, \dots, x_n]/I_f$

is zero in sufficiently high dimensions. Thus it has trivial ${\mathrm{Proj}}$. And the fiber over ${D(f)}$ is empty, completing the proof.

Anyway, what’s the moral of all this? One is that properness and separatedness don’t require noetherian hypotheses to work, and the “non-elementary” means of valuative criteria is unnecessary to prove the properness of projective morphisms.

Incidentally, properness is not synonymous with projectivity. Under reasonable hypotheses, proper morphisms can be approximated by projective morphisms, though; this is called Chow’s lemma, and is crucial for proving things like the direct image theorem (the push-forward of a coherent sheaf by a proper map is coherent–ditto for the higher direct images).