# 1 Separated and proper morphisms

## The functorial point of view

**(1.1) Schemes as functors**

April 3,

2023

References: [ GW1 ] Sections (4.1), (4.2); [ Mu ] II.6.

As we have discussed in Algebraic Geometry 1, to a scheme \(X\) we can attach its functor of \(T\)-valued points:

which on morphisms is just given by composition: \(f\colon T'\to T\) is mapped under \(h_X\) to the map \(X(T)\to X(T')\), \(\alpha \mapsto \alpha \circ f\).

**Example 1.1**

For any affine scheme \(X=\operatorname{Spec}A\), we have \(X(T) = \operatorname{Hom}(A, \Gamma (T, {\mathscr O}_T))\). For example, this gives \({\mathbb A}^n_R(T) = \Gamma (T, {\mathscr O}_T)^n\) for any ring \(R\) and any \(R\)-scheme \(T\) (where we understand \({\mathbb A}^n_R(T)\) as the set of morphisms \(T\to {\mathbb A}^n_R\)

*of \(R\)-schemes*).It is more difficult to describe \(\mathbb {P}^n(T)\) for general \(T\) (for \(T\) the spectrum of a field, we have the description by homogeneous coordinates). We will come back to this later.

Using the notion of a morphism of functors, we can speak of the category \(\widehat{{\mathscr C}} := \operatorname{Func}(({\rm Sch})^{{\rm opp}}, {\rm (Sets)})\) of all such functors, and we obtain a functor \(h\colon ({\rm Sch}) \to \widehat{{\mathscr C}}\), \(X\mapsto h_X\), which on morphisms is – once again – defined by composition: \(\alpha \colon X'\to X\) is mapped by \(h\) to the morphism \(h_{X'}\to h_X\) of functors given by \(X'(T)\to X(T)\), \(\beta \mapsto \beta \circ \alpha \).

Even though at first sight this may look complicated, this is an entirely “formal” (i.e., category-theoretic) procedure which has nothing to do with schemes. In fact, if \({\mathscr C}\) is any category, for an object \(X\) of \({\mathscr C}\) we can define the functor

and now setting \(\widehat{{\mathscr C}} := \operatorname{Func}({\mathscr C}^{{\rm opp}}, {\rm (Sets)})\), we obtain a functor \(h\colon {\mathscr C}\to \widehat{{\mathscr C}}\).

**Theorem 1.2**

Given \(X\) and \(Y\) and a morphism \(\Phi \colon h_X\to h_Y\), we obtain a morphism \(X\to Y\) by applying \(\Phi \) to \(\operatorname{id}_X\in h_X(X)\). One checks that this is an inverse of the map \(\operatorname{Hom}(X, Y)\to \operatorname{Hom}(h_X, h_Y)\) given by \(h\).

**Remark 1.3**

We will mostly apply the Yoneda Lemma to the category of schemes, or the category of \(S\)-schemes for some fixed scheme \(S\). Let us list some of its consequences (as before, these facts are not specific to the category of schemes):

Let \(X\), \(Y\) be schemes. The following are equivalent:

\(X\cong Y\),

\(h_X \cong h_Y\) (isomorphism of functors)

there exists a family \(f_T\colon X(T)\to Y(T)\) of bijections of sets that is functorial in \(T\), i.e., for every scheme morphism \(T'\to T\), the diagram

is commutative.

Let \(X\), \(Y\) be schemes. Giving a scheme morphism \(X\to Y\) is equivalent to giving a family of maps \(f_T\colon X(T)\to Y(T)\) of sets for each scheme \(T\), that is functorial in \(T\) (same condition as in (1) (iii)).

*Example.*The determinant of a matrix is a scheme morphism \(\mathbb A^{n^2}\to \mathbb A^1\).A diagram of scheme morphisms is commutative if and only if for every scheme \(T\) the diagram (in the category of sets) obtained by replacing each scheme by its set of \(T\)-valued points, and replacing the scheme morphisms by the induced maps of sets, is commutative.

## Fiber products, base change, and separated morphisms

**(1.2) Recap: fiber products of schemes**

April 5,

2023

References: [ GW1 ] Sections (4.4)–(4.6); [ H ] II.3; [ Mu ] II.2.

The universal property of a *fiber product* generalizes the universal property of a product (of two objects, in any category). It is defined as follows. (See the lecture notes for Algebraic Geometry 1 for a bit more background, e.g., an explanation of the term *fiber product*.)

**Definition 1.4**

*fiber product*of \(X\) and \(Y\) over \(S\), if for every object \(T\) of \({\mathscr C}\) together with morphisms \(\alpha \colon T\to X\) and \(\beta \colon T\to Y\) with \(f\circ \alpha = g\circ \beta \), there exists a unique morphism \(\xi \colon T\to P\) with \(p\circ \xi = \alpha \), \(q\circ \xi = \beta \).

We say that a commutative square

is *cartesian*, if it is a fiber product diagram, i.e., if \(A\) satisfies the universal property defining the fiber product of \(B\) and \(C\) over \(D\).

**Theorem 1.5**

If \(f\colon X\to S\), \(g\colon Y\to S\) are morphisms of schemes, then the fiber product of \(X\) and \(Y\) over \(S\) exists.

If in (1) \(X=\operatorname{Spec}A\), \(Y=\operatorname{Spec}B\), \(S=\operatorname{Spec}R\) are affine schemes (so that \(f\) and \(g\) are given by ring homomorphisms \(R\to A\), \(R\to B\), then \(\operatorname{Spec}A\otimes _RB\) together with the morphisms induced by the natural maps \(A\to A\otimes _RB\), \(B\to A\otimes _RB\), is the fiber product of \(X\) and \(Y\) over \(S\).

For \(f\), \(g\) as in (1), and open covers \(S = \bigcup _i U_i\), \(f^{-1}(U_i) = \bigcup _j V_{ij}\), \(g^{-1}(U_i) = \bigcup _k W_{ik}\), for all \(i\), \(j\), \(k\), the natural morphism \(V_{ij}\times _{U_i}W_{ik}\to X\times _SY\) induced by the universal property of the fiber product is an open immersion, and taken together the open subschemes of the above form cover \(X\times _SY\).

**Example 1.6**

**Example 1.7**

*scheme-theoretic fiber of \(f\) over \(s\)*. The projection \(X_s\to X\) induces a homeomorphism between the topological space of the scheme \(X_s\) and the fiber of the continuous map \(f\) over \(s\). Cf. the Algebraic Geometry 1 class, or see [ GW1 ] Section (4.8).

**Lemma 1.8**

\(X\times _SS\cong X\),

\(X\times _SY \cong Y\times _SX\),

\((X\times _SY)\times _TZ \cong X\times _S(Y\times _TZ)\) (and this allows us to omit the parentheses in expressions like these).

These properties can easily be checked using the universal property (or, what more or less amounts to the same, by the Yoneda lemma). In any case, this reduces to checking the above claims for fiber products of sets, where they follow immediately from the explicit description of fiber products of sets.

April 12,

2023

**Example 1.9**

(Group schemes) Let \(S\) be a scheme. A *group scheme over \(S\)* is an \(S\)-scheme \(G\) together with a functor \(h\colon ({\rm (Sch)}/S)^{{\rm opp}}\to {\rm (Grp)}\), such that \(h_G\) is the composition of \(h\) and the forgetful functor \({\rm (Grp)}\to {\rm (Sets)}\). In other words, for every \(S\)-scheme \(T\), we are given a group structure on \(G(T)\), and for every morphism \(T'\to T\), the induced map \(G(T)\to G(T')\) is a group homomorphism.

In view of the above discussion, we can express this structure equivalently by giving a multiplication morphism \(m\colon G\times _SG\to G\), a morphism \(i\colon G\to G\) (“inverse element”) that induces the map \(g\mapsto g^{-1}\) on each \(G(T)\), and a morphism \(S\to G\) (“neutral element”) that induces the neutral element in each \(G(T)\) (note that for every \(S\)-scheme \(T\), the set \(S(T)\) is a singleton). The morphisms \(m\), \(i\), \(e\) have to satisfy certain conditions reflecting the group axioms; the conditions can be expressed by requiring that certain diagrams be commutative. See [ GW1 ] Section (4.15).

**(1.3) Base change**

References: [ GW1 ] Chapter 4, in particular Sections (4.7)–(4.10).

Given scheme morphisms \(f\colon X\to S\) and \(g\colon S'\to S\), we call the projection \(X\times _SS'\to S'\) the *morphism obtained from \(f\) by base change along \(g\)*. This defines a functor from the category of \(S\)-schemes to the category of \(S'\)-schemes.

A particularly simple example is the case where \(g\colon V\to S\) is an open immersion. In that case the base change of \(f\) is just the restriction of \(f\) to \(f^{-1}(V)\to V\).

Many properties of scheme morphisms are “stable under base change” in the following sense: A property \(\mathbf P\) of scheme morphisms is called stable under base change if for every morphism \(f\colon X\to Y\) of \(S\)-schemes that has property \(\mathbf P\) and every scheme morphism \(S'\to S\), the induced morphism \(X\times _SS'\to Y\times _SS'\) also has property \(\mathbf P\).

Given a property \(\mathbf P\), to check that it is stable under base change, it is enough to check that whenever \(f\colon X\to S\) has the property, and \(g\colon S'\to S\) is a scheme morphism, then \(X\times _SS'\to S'\) also has the property. In fact, this is clearly a special case of the above definition (namely the case where \(Y=S\)). On the other hand, suppose this special case is true and \(f\colon X\to Y\) is any morphism of \(S\)-schemes. Identifying \(X\times _SS' = X\times _Y(Y\times _SS')\) using the rules of “computations with fiber products” (Lemma 1.8), the base change \(X\times _SS'\to Y\times _SS'\) is identified with the projection \(X\times _Y(Y\times _SS')\to Y\times _SS'\). Applying the special case to \(X\to Y\) and the base change \(Y\times _SS'\to Y\), we obtain that \(X\times _SS' = X\times _Y(Y\times _SS')\to Y\times _SS'\) has property \(\mathbf P\).

**Proposition 1.10**

an open immersion,

a closed immersion,

an immersion,

quasi-compact,

surjective,

an isomorphism,

*…most of the properties of scheme morphisms that we will get to know later in the course …*

A notable exception is the property of being injective: Can you find an example of an injective morphism \(X\to S\) of schemes and a morphism \(S'\to S\) such that the base change \(X\times _SS'\to S'\) is not injective?

All the properties in the above list, and also being injective, are *stable under composition*, i.e., if two composable morphisms both have the property, then so does the composition.

**Example 1.11**

If \(R\to R'\) is a ring homomorphism, then \(\mathbb {A}^n_R\otimes _RR' := \mathbb {A}^n_R\times _{\operatorname{Spec}R}\operatorname{Spec}R' = \mathbb {A}^n_{R'}\). In view of this we define, for an arbitrary scheme \(S\), \(\mathbb {A}^n_S := \mathbb {A}^n_{\mathbb {Z}}\times _{\operatorname{Spec}\mathbb {Z}} S\).

If \(R\to R'\) is a ring homomorphism, then \(\mathbb {P}^n_R\otimes _RR' := \mathbb {P}^n_R\times _{\operatorname{Spec}R}\operatorname{Spec}R' = \mathbb {P}^n_{R'}\). In view of this we define, for an arbitrary scheme \(S\), \(\mathbb {P}^n_S := \mathbb {P}^n_{\mathbb {Z}}\times _{\operatorname{Spec}\mathbb {Z}} S\).

**(1.4) Separated morphisms**

References: [ GW1 ] Sections (9.3), (9.4); [ H ] II.4.

Recall the following description of a topological space \(X\) being Hausdorff: \(X\) is Hausdorff if and only if the “diagonal” \(\{ (x,x)\in X\times X;\ x\in X\} \) is a *closed* subset of \(X\times X\) (with respect to the product topology).

As we have discussed, the underlying topological space of a scheme usually is not Hausdorff. On the other hand, the (fiber) product of schemes usually does not carry the product topology, and it turns out that using the above condition in the context of schemes gives rise to an interesting and useful notion for schemes (which in some sense is a good replacement for the Hausdorff property; see Proposition 1.17 for an example).

**Definition 1.12**

*separated*\(S\)-scheme, or that \(X\) is separated over \(S\), or that the morphism \(X\to S\) is separated, if \(\Delta \) is a closed immersion.

April 17,

2023

**Remark 1.13**

Suppose that \(X\to S\) is a morphism of affine schemes, corresponding to a ring homomorphism \(\varphi \colon A\to B\), say. Then \(X\times _SX\) also is affine and the diagonal \(X\to X\times _SX\) corresponds to the ring homomorphism \(B\otimes _AB\to B\), \(b\otimes b'\mapsto bb'\). Since this ring homomorphism is clearly surjective, \(\Delta \) is a closed immersion. Thus we see that every morphism between affine schemes is separated.

For every morphism \(X\to S\), the diagonal \(\Delta \colon X\to X\times _SX\) is an immersion. Therefore we could phrase the condition that \(X\) is separated over \(S\) equivalently by requiring that the image \(\Delta (X)\) is a closed subset of (the topological space of the scheme) \(X\times _SX\).

Let \(k\) be a field, and let \(X\) be the “affine line with doubled origin”. Then \(X\) is not separated over \(k\). (The “valuative criterion for separatedness” shows that this is really a typical example of a non-separated scheme.)

Note that the notion of being

*separated*has nothing to do with the notion of being*separable*(as in separable field extension; there also exists the notion of separable scheme morphism).

**Proposition 1.14**

The morphism \(f\) is separated.

For all affine open subschemes \(U, V\subseteq X\), the intersection \(U\cap V\) is affine and the natural homomorphism

\[ \Gamma (U, {\mathscr O}_X)\otimes _R\Gamma (V,{\mathscr O}_X)\to \Gamma (U\cap V, {\mathscr O}_X) \]is surjective.

There exists an affine open cover \(X=\bigcup _i U_i\) such that for all \(i, j\) the intersection \(U_i\cap U_j\) is affine and the natural homomorphism

\[ \Gamma (U_i, {\mathscr O}_X)\otimes _R\Gamma (U_j,{\mathscr O}_X)\to \Gamma (U_i\cap U_j, {\mathscr O}_X) \]is surjective.

The key point is noting that for \(U, V\subseteq X\) open, \(U\times _XV\) can be identified with \(U\cap V\), and that as a consequence the diagram

is cartesian.

**Corollary 1.15**

**Corollary 1.16**

The following proposition illustrates (why?) that in some sense the property of being separated resembles the Hausdorff property of topological spaces. (See Algebraic Geometry 1, Problem 52, for an example which shows that the reducedness assumption cannot be dropped.)

**Proposition 1.17**

**Remark 1.18**

Every immersion and more generally every monomorphism of schemes is separated. (A morphism \(X\to Y\) of schemes is called a monomorphism, if for every scheme \(T\) the map \(X(T)\to Y(T)\) is injective. This is the “usual” categorical notion of monomorphism. Every immersion is a monomorphism. If \(X\to S\) is a monomorphism, then the diagonal morphism \(X\to X\times _SX\) is an isomorphism, and a fortiori a closed immersion.)

The composition of separated morphisms is separated. The base change of a separated morphism is separated.

The property of being separated can be checked locally on the target.

Given a morphism \(f\colon X\to Y\), \(f\) is separated if and only if the induced morphism \(f_{\operatorname{red}}\) between the underlying reduced subschemes of \(X\) and \(Y\) is separated.

## Proper morphisms

**(1.5) Proper maps between topological spaces**

References: [ Bou-TG ] Ch. I §10.

Most schemes that we have encountered so far (in particular, all affine schemes, projective space over any ring, subschemes \(V_+(I)\) of projective space over a ring, …) are quasi-compact. On the other hand, from a geometric point of view, e.g., the affine line (or higher-dimensional affine space) “should not be viewed” as a compact space. The notion of properness is a suitable replacement in algebraic geometry for the notion of compactness in topology/differential geometry.

Similarly as separatedness, we will define properness in terms of fiber products of schemes, starting from a characterization of quasi-compact topological spaces, given by the notion of *proper map* between continuous spaces, which we discuss below as a motivation for the definition of proper scheme morphisms. The purpose of motivation aside, the rest of this section plays no role in the course.

Note that fiber products in the category of topological spaces exist. In fact, for continous maps \(X\to S\), \(Y\to S\), the set-theoretic fiber product \(X\times _SY\), equipped with the subspace topology for the inclusion \(X\times _SY\subseteq X\times Y\) (where the right hand side carries the product topology) is easily seen to satisfy the required universal property.

Recall that a continous map \(f\colon X\times Y\) is *closed*, if for every closed subset \(C\subseteq X\), the image \(f(C)\subseteq Y\) is closed.

**Definition 1.19**

We call a continuous map \(f\colon X\to Y\)

*universally closed*, if for every continuous map \(Z\to Y\), the “base change” of \(f\) along \(Z\to Y\), i.e., the induced map \(X\times _YZ\to Z\), is closed.We call a continuous map \(f\colon X\to Y\)

*proper*, if for every topological space \(Z\), the induced map \(f\times \operatorname{id}_Z\colon X\times Z\to Y\times Z\) is closed.

**Proposition 1.20**

Let \(f\colon X\to Y\) be a continuous map. Consider the following properties.

\(f\) is universally closed,

\(f\) is proper,

\(f\) is closed and for every \(y\in Y\) the fiber \(f^{-1}(y)\) is quasi-compact.

for every quasi-compact subset \(K\subseteq Y\), the inverse image \(f^{-1}(K)\) is quasi-compact.

Then (i) \(\Rightarrow \) (ii) \(\Leftrightarrow \) (iii) \(\Rightarrow \) (iv).

If in addition \(Y\) is Hausdorff, then (i) \(\Leftrightarrow \) (ii).

If in addition \(Y\) is Hausdorff and locally compact, then all the above properties are equivalent.

In the remainder of the section, we will discuss the proof of the proposition. It is easy to see that (i) implies (ii). In fact, if \(Z\) is a topological space, then \(X\times Z = X\times _Y(Y\times Z)\), so the map occurring in (ii) can be written as the base change of \(f\) along the projection \(Y\times Z\to Y\).

Conversely, if \(Y\) is Hausdorff, and \(g\colon Z\to Y\) is continuous, then \(X\times _YZ\) is closed in \(X\times Z\). Similarly the map \(Z\to Y\times Z\), \(z\mapsto (g(z), z)\), which up to switching the factors is the graph of \(g\), identifies \(Z\) with a closed subset of \(Y\times Z\). Now let \(C\subseteq X\times _YZ\) be closed, and write \(f_Z\colon X\times _YZ\to Z\) for the base change of \(f\). Then \(f_Z(C)\subseteq Z\), under the inclusion \(Z\hookrightarrow Y\times Z\), is identified with the image of \(C\) under the composition \(X\times _YZ\to X\times Z\to Y\times Z\). This shows that (ii) implies (i), if \(Y\) is Hausdorff.

Denote by \(P\) the topological space consisting of a single point.

In the following lemma, which in a sense is the most interesting part of the proposition, we consider the special case \(Y=P\). In this case, properties (i) and (ii) are equivalent for trivial reasons (since \(P\) is a terminal object in the category of topological spaces, fiber products over \(P\) are just products). Likewise, (iii) and (iv) are equivalent in this case. The lemma proves that, for \(Y=P\), conditions (i) and (ii) imply conditions (iii) and (iv).

**Lemma 1.21**

We need to show that given any cover of \(X\) by open subsets, finitely many of these suffice to cover \(X\). Passing to complements, we can equivalently show the following statement: Let \((Z_i)_{i\in I}\) be a family of closed subsets of \(X\) such that for every finite subset \(J\subseteq I\) the intersection \(\bigcap _{i\in J} Z_i\) is non-empty. Then the intersection \(\bigcap _{i\in I} Z_i\) is non-empty.

So take a family \((Z_i)_i\) of closed subsets of \(X\) such that all finite intersections are non-empty. In particular, all the \(Z_i\) are themselves non-empty. Adding further sets to this family, we may assume for every finite subset \(J\subseteq U\), the intersection \(\bigcap _{i\in J} Z_i\) is itself one of the closed subsets in this family. We will use that the map \(X\times X'\to X'\) is closed for the following topological space \(X'\).

As a set \(X'\) is the disjoint union of \(X\) and a set consisting of one element \(\omega \), i.e., we formally add one further point to \(X\). The topology on \(X'\) is defined as follows. (Note that the topology on \(X\) is *not* the subspace topology for the inclusion \(X\to X'\) in most cases; in fact, this inclusion will not be continuous unless \(X\) carries the discrete topology.) The open sets of \(X'\) are the sets that can be obtained as a union of sets of the form \(\{ x\} \), \(x\in X\), and \(Z_i\cup \{ \omega \} \), \(i\in I\). One checks that this defines a topology, i.e., that \(\emptyset \) and \(X'\) are open, and that arbitrary unions and finite intersections of such sets again have this form. Only for the third point there is really something to be checked, but the claim easily follows in view of our assumptions on the family \((Z_i)_i\).

Note that the closure of the subset \(X\subseteq X'\) in \(X'\) is all of \(X'\) since every open subset that contains \(\omega \) meets \(X\).

As indicated above, we will use that the projection \(p_2\colon X\times X'\to X'\) is closed by our assumption that the map \(X\to P\) is proper. Denoting by \(C\) the closure of the subset \(\Delta :=\{ (x,x);\ x\in X\} \subseteq X\times X'\), we obtain that \(p_2(C)\) is closed. Since it clearly contains all points of \(X\subseteq X'\), by the above remark we see that \(\omega \in p_2(C)\). This means that there exists \(\xi \in X\) such that \((\xi , \omega )\) lies in the closure of \(\Delta \) (with respect to the product topology).

Let us show that \(\xi \in \bigcap _{i\in I} Z_i\). Saying that \((\xi , \omega )\) lies in the closure of \(\Delta \) means that every neighborhood of \((\xi , \omega )\) in \(X\times X'\) meets \(\Delta \). In particular, for every neighborhood \(U\subseteq X\) of \(\xi \) and every \(i\), \(U\times (Z_i\cup \{ \omega \} )\) meets \(\Delta \). Fixing \(i\), we see that \(Z_i\) meets every neighborhood of \(\xi \) in \(X\), i.e., \(\xi \) lies in the closure of \(Z_i\). Since \(Z_i\) is closed, this means \(\xi \in Z_i\), as we wanted to show.

It is easy to see that whenever \(f\colon X\to Y\) satisfies (ii) and \(Y'\subseteq Y\) is a subspace, then the restriction \(f^{-1}(Y')\to Y'\) of \(f\) satisfies (ii) as well. Applying this to singletons \(Y'\), from the lemma we obtain the implication (ii) \(\Rightarrow \) (iii) in the proposition.

For the implication (iii) \(\Rightarrow \) (iv), to ease the notation we first note that (replacing \(Y\) by \(K\) and \(X\) by \(f^{-1}(K)\) and observing that (iii) still holds) it is enough to handle the case that \(K=Y\) is quasi-compact. We then have to show that \(X\) is quasi-compact. So let \(X = \bigcup U_i\) be an open cover of \(X\). By quasi-compactness of the fibers of \(f\), for each \(y\) there exists an open \(U_y\subseteq X\) containing \(f^{-1}(y)\) which is a finite union of subsets of the form \(U_i\). Since \(f\) is closed, \(V_y:=Y\setminus f(X\setminus U_y)\) is open in \(Y\). It clearly contains \(y\), so all the \(V_y\) together cover \(Y\). By quasi-compactness of \(Y\), finitely many \(V_y\) suffice to cover \(Y\). But then the corresponding \(U_y\) cover \(X\), and we are done.

Let us show that (iii) implies (ii), so let \(f\colon X\to Y\) as in (iii) be given and let \(Z\) be any topological space. Let \(C\subseteq X\times Z\) be closed. We show that \(W:=(Y\times Z)\setminus (f\times \operatorname{id}_Z)(C)\) is open in \(Y\times Z\). For \((y,z)\in W\), we need to find open neighborhoods \(U\) of \(y\) in \(Y\) and \(V\) of \(z\) in \(Z\) such that \(U\times V \subseteq W\), or in other words: \((f^{-1}(U)\times V)\cap C=\emptyset \). If \(y\) does not lie in the image of \(f\), we can take \(U\) to be any neighborhood of \(y\) which is disjoint from the image of \(f\) (which exists since \(f\) is closed). Otherwise, for each \(x\in f^{-1}(y)\), let \(U_x\) be an open neighborhood of \(x\) in \(X\) and \(V_x\) an open neighborhood of \(z\) in \(Z\) auch that \((U_x\times V_x)\cap C = \emptyset \). Since \(f^{-1}(y)\) is quasi-compact, finitely many of the \(U_x\) suffice to cover this fiber. Let \(U'\) denote their union, and \(V\) the intersection of the corresponding sets \(V_x\). Since \(f\) is closed, \(U:= Y\setminus f(X\setminus U')\) is an open neighborhood of \(y\). One checks that \(U\) and \(V\) constructed in this way have the desired property.

Finally, let us discuss the implication (iv) \(\Rightarrow \) (iii). As we will see, we will have to assume that \(Y\) is Hausdorff and that every \(y\) admits a quasi-compact neighborhood, i.e., there exist an open \(U\) and a quasi-compact \(K\subseteq Y\) such that \(y\in U\subseteq K\). Clearly, conditions (iv) implies that all fibers of \(f\) are quasi-compact (without any assumption on \(Y\)). We need to show that \(f\) is closed.

Let \(C\subseteq X\) be a closed subset. To check that \(f(C)\) is closed, we may work locally on \(Y\). Since property (iv) still holds if we replace \(Y\) by a subspace (and \(X\) by the inverse image of that subspace), and since \(Y\) is locally compact, we may assume without loss of generality that \(Y\) is quasi-compact, and — using our assumption (iv) — that \(X\) is quasi-compact as well. Then \(C\) is also quasi-compact, being closed in a quasi-compact space, and thus the image \(f(C)\) is quasi-compact, as well. Since \(Y\) is Hausdorff and quasi-compact subspaces of a Hausdorff space are necessarily closed, we see that \(f(C)\) is closed in \(Y\), as desired.

**(1.6) Proper morphisms**

References: [ GW1 ] Sections (12.13); [ H ] II.4; [ Mu ] II.7.

April 19,

2023

To define proper morphisms of schemes, we also need the following ingredients. We have already defined quasi-compact morphisms, and a special case of morphisms of finite type in Algebraic Geometry 1.

**Definition 1.22**

*quasi-compact*, if for every quasi-compact open \(V\subseteq Y\) the inverse image \(f^{-1}(V)\) is quasi-compact.

**Lemma 1.23**

The morphism \(f\) is quasi-compact.

For every affine open subscheme \(V\subseteq Y\), the inverse image \(f^{-1}(V)\) is quasi-compact.

There exists a cover \(Y=\bigcup _i V_i\) by affine open subschemes such that for every \(i\) the inverse image \(f^{-1}(V_i)\) is quasi-compact.

The difficult implication was Problem 46 in Algebraic Geometry 1. Note that in part (iii) of the lemma it is important to consider a cover by *affine* open subschemes.

Recall that an algebra \(B\) over a ring \(A\) is called *of finite type* (or equivalently, *finitely generated*) if there exists \(n\ge 0\) and a surjective \(A\)-algebra homomorphism \(A[X_1,\dots , X_n]\to B\).

**Definition 1.24**

A morphism \(f\colon X\to Y\) of schemes is called

*locally of finite type*(or: \(X\) is called a \(Y\)-scheme locally of finite type, of locally of finite type over \(Y\)), if for every affine open subscheme \(V\subseteq Y\) and every open subscheme \(U\subseteq f^{-1}(V)\), the ring homomorphism \(\Gamma (V, {\mathscr O}_Y)\to \Gamma (U, {\mathscr O}_X)\) induced by the restriction \(U\to V\) of \(f\) makes \(\Gamma (U, {\mathscr O}_X)\) a \(\Gamma (V, {\mathscr O}_Y)\)-algebra of finite type.A morphism \(f\colon X\to Y\) of schemes is called

*of finite type*(or: \(X\) is called a \(Y\)-scheme of finite type, or of finite type over \(Y\)), if \(f\) is locally of finite type and quasi-compact.

**Lemma 1.25**

The morphism \(f\) is locally of finite type.

There exist a cover \(Y=\bigcup _i V_i\) by affine open subschemes, and for each \(i\) a cover \(f^{-1}(V_i) = \bigcup _j U_{ij}\) by affine open subschemes such that for all \(i\), \(j\) the \(\Gamma (V_i, {\mathscr O}_Y)\)-algebra \(\Gamma (U_{ij}, {\mathscr O}_X)\) is of finite type.

Each of the properties of being *locally of finite type*, *quasi-compact*, and *of finite type* is stable under composition and under base change.

**Definition 1.26**

The morphism \(f\) is called

*closed*, if for every closed subset \(Z\subseteq X\), the image \(f(Z)\) is a closed subset of \(Y\).The morphism \(f\) is called

*universally closed*, if for every morphism \(Y'\to Y\) the base change \(X\times _YY'\to Y'\) of \(f\) along \(Y'\) is a closed morphism.The morphism \(f\) is called

*proper*, if it is separated, of finite type, and universally closed.

**Example 1.27**

The affine line is not proper. More precisely, let \(k\) be a field, let \(Y=\operatorname{Spec}(k)\), and let \(X = \mathbb {A}^1_k\). Let \(f\colon X\to Y\) be the natural morphism. Then \(f\) is separated, of finite type and closed, but (why?) not universally closed.

Every closed immersion is proper.

The property of being proper is stable under composition and under base change.

**(1.7) Projective schemes are proper**

**Definition 1.28**

*projective*(we also say that \(X\) is projective over \(S\), or that the morphism \(X\to S\) is projective), if there exist \(N\ge 0\) and a closed immersion \(X\hookrightarrow \mathbb {P}^N_S\) of \(S\)-schemes.

This definition of *projective schemes* differs slightly from the one in
[
GW1
]
(Definition 13.68, which requires only that the above property holds locally on \(S\)). If \(S\) is affine, they coincide, however, and the difference will not be of any concern for us in this course. See
[
GW1
]
, Summary 13.71 for a discussion. The definition given here is the one used in
[
H
]
and in
[
Stacks
]
.

Suppose that \(S= \operatorname{Spec}R\) is an affine scheme. For any homogeneous ideal \(I\subseteq R[X_0,\dots , X_N]\), \(V_+(I)\) is a closed subscheme of \(\mathbb {P}^N_S\), and hence in particular a projective \(S\)-scheme. One can show that for \(S\) affine every projective scheme is isomorphic to a scheme of this form.

We will study this notion in more detail later (see Chapter 5).

Before we come to the main theorem of this section (Theorem 1.32), recall that for a homogeneous ideal \(I\subseteq R[X_0,\dots , X_n]\) (where \(R\) is some ring) we have defined a closed subscheme \(V_+(I)\) of \(\mathbb {P}^n_R\). We need the following two results on closed subschemes of projective space.

**Lemma 1.29**

**Proposition 1.30**

In addition, we will use the following commutative algebra lemma which is easily proved using the definitions of the localizations appearing in the lemma. (We will later generalize the lemma when we prove that given an \({\mathscr O}_X\)-module \({\mathscr F}\) of finite type on a locally ringed space \(X\), the support of \({\mathscr F}\), i.e., the set of all points \(x\) such that the stalk \({\mathscr F}_x\) does not vanish, is closed. See Proposition 2.18.)

**Lemma 1.31**

**Theorem 1.32**

Since closed immersions are proper, it is enough to prove that projective space is proper, i.e., that for every scheme \(S\) the morphism \(\mathbb {P}^n_S\to S\) is closed. Since this property can be checked locally on \(S\), we may assume that \(S=\operatorname{Spec}R\) is affine.

If \(Z\subseteq \mathbb {P}^n_S\) is a closed subset, there exists a closed subscheme with underlying topological space \(Z\), and hence (Proposition 1.30) a homogeneous ideal \(I\subseteq R[X_0,\dots , X_n]\) such that \(V_+(I)\) has underlying topological space \(Z\). We need to show that the image of \(V_+(I)\) in \(S\) is closed, or equivalently, that its complement \(U\subseteq S\) is open.

Denote by \(f\) the composition \(V_+(I)\hookrightarrow \mathbb {P}^n_S\to S\), and let \(x\in U\). Then the scheme-theoretic fiber \(f^{-1}(x) = V_+(I) \times _U\operatorname{Spec}\kappa (x)\) is empty. We want to show that there exists \(s\in R\) such that \(x\in D(s)\subseteq U\). The inclusion \(D(s)\subseteq U\) amounts to saying that \(f^{-1}(D(s)) = \emptyset \). To translate the problem into a commutative algebra statement, let \(\overline{I}\) be the image of \(I\) in \(\kappa (x)[X_0,\dots , X_n]\). It follows from the assumption \(f^{-1}(x) = \emptyset \) and Lemma 1.29 that \(\operatorname{rad}(\overline{I})\) contains the ideal \((X_0, \dots , X_n)\) (\(\subset \kappa (x)[X_0, \dots , X_n]\)). Thus for \(d\) sufficiently large, for the degree \(d\) components we have \(\overline{I}_d = (X_0,\dots , X_n)_d\). By the lemma of Nakayama, we obtain \(I_d\otimes {\mathscr O}_{S, x} = (X_0, \dots , X_n)_d \subseteq {\mathscr O}_{S,x}[X_0, \dots , X_n]\). It then follows that the analogous equality holds already over the localization of \(R\) with respect to a suitable element \(s\) not contained in the prime ideal \(x\).

April 24,

2023

**Example 1.33**

(The resultant of polynomials) Let \(k\) be an algebraically closed field (with some “obvious” adaptations, the results below hold over an arbitrary field). Let \(m,n\in \mathbb {N}\). We identify the set \(A\) of pairs \((f,g)\) of monic polynomials with the set of \(k\)-valued points of the affine space \(\mathbb {A}^{m+n}_k = \operatorname{Spec}k[S_0,\dots , S_{m-1}, T_0,\dots , T_{n-1}]\), where a tuple \((s_i, t_j)\in k^{m+n} = \mathbb {A}^{m+n}_k(k)\) corresponds to \(\left(X^m+s_{m-1}X^{m-1}+\cdots + s_0, X^n+t_{n-1}X^{n-1}+\cdots + t_0 \right)\).

Viewing \(A\) as the set of closed points of \(\mathbb {A}^{m+n}_k\), \(A\) is equipped with a topology, namely the topology induced by the Zariski topology.

Let \(Z\subset A\) be the subset consisting of those pairs \((f,g)\) such that \(f\) and \(g\) have a common zero in \(k\). Write \(R=k[S_0,\dots , S_{m-1}, T_0,\dots , T_{n-1}]\).

*Claim.* The set \(Z\) is a closed subset.

*Proof of claim.* Let

be the “universal” monic polynomials, and let

be their homogenizations with respect to a second variable \(Y\).

Let \(p\colon \mathbb {P}^1_R\to \operatorname{Spec}R\) be the projection. Then \(Z = p(V_+(\tilde{F}, \tilde{G}))\cap A\). By the above theorem, \(p(V_+(\tilde{F}, \tilde{G}))\) is closed in \(\mathbb {A}^{m+n}_k\), hence the claim follows. To see the equality, fix \(x = (f,g)\in A\) and let \(\tilde{f}, \tilde{g}\) be their homogenizations. Then \(f\) and \(g\) have a common zero in \(k\) if and only if \(V_+(\tilde{f}, \tilde{g}) \ne \emptyset \) (inside \(\mathbb {P}^1_{\kappa (x)}\)). Note that the point \((1:0)\), the “point at infinity” in \(\mathbb {P}^1\) is never a zero of \(\tilde{f}\) or \(\tilde{g}\). Since \(V_+(\tilde{f}, \tilde{g}) = V_+(\tilde{F}, \tilde{G})\times _{\operatorname{Spec}R}\operatorname{Spec}\kappa (x)\) can be identified with the (scheme-theoretic) fiber of \(p\) over the point \(x\), this proves the desired description of \(Z\).

More precisely one can show (using other methods) that \(Z\) is the zero locus of a single polynomial in \(R\), the so-called *resultant* of a pair of monic polynomials. See
[
GW1
]
Section (B.20) for a sketch and further references, or
[
Bo
]
Abschnitt 4.4 for a detailed account in German.