## II.3: First Properties of Schemes

### II.3.1

Show that a morphism $f: X \rightarrow Y$ is locally of finite type if and only if for every open affine subset $V=\operatorname{Spec} B$ of $Y, f^{-1}(V)$ can be covered by open affine subsets $U_j=\operatorname{Spec} A_j$, where each $A_j$ is a finitely generated $B$-algebra.

### II.3.2

A morphism $f: X \rightarrow Y$ of schemes is **quasi-compact** if there is a cover of $Y$ by open affines $V_i$ such that $f^{-1}\left(V_i\right)$ is quasi-compact for each $i$. 

Show that $f$ is quasicompact if and only if for every open affine subset $V \subseteq Y, f^{-1}(V)$ is quasi-compact.

### II.3.3

a. Show that a morphism $f: X \rightarrow Y$ is of finite type if and only if it is locally of finite type and quasi-compact.

b. Conclude from this that $f$ is of finite type if and only if for every open affine subset $V=\operatorname{Spec} B$ of $Y, f^{-1}(V)$ can be covered by a finite number of open affines $U_j=\operatorname{Spec} A_j$, where each $A_j$ is a finitely generated $B$-algebra.

c. Show also if $f$ is of finite type, then for every open affine subset $V=\operatorname{Spec} B \subseteq$ $Y$, and for every open affine subset $U=\operatorname{Spec} A \subseteq f^{-1}(V), A$ is a finitely generated $B$-algebra.

### II.3.4

Show that a morphism $f: X \rightarrow Y$ is finite if and only if for every open affine subset $V=\operatorname{Spec} B$ of $Y, f^{-1}(V)$ is affine, equal to $\operatorname{Spec} A$, where $A$ is a finite $B$-module.

### II.3.5

A morphism $f: X \rightarrow Y$ is **quasi-finite** if for every point $y \in Y, f^{-1}(y)$ is a finite set.

a. Show that a finite morphism is quasi-finite.

b. Show that a finite morphism is closed, i.e., the image of any closed subset is closed.

c. Show by example that a surjective, finite-type, quasi-finite morphism need not be finite.

### II.3.6

Let $X$ be an integral scheme. Show that the local ring $\mathcal{O}_{\xi}$ of the generic point $\xi$ of $X$ is a field. It is called the **function field** of $X$, and is denoted by $K(X)$. 

Show also that if $U=\operatorname{Spec} A$ is any open affine subset of $X$, then $K(X)$ is isomorphic to the quotient field of $A$.

### II.3.7

A morphism $f: X \rightarrow Y$, with $Y$ irreducible, is **generically finite** if $f^{-1}(\eta)$ is a finite set, where $\eta$ is the generic point of $Y$. A morphism $f: X \rightarrow Y$ is **dominant** if $f(X)$ is dense in $Y$. 

Now let $f: X \rightarrow Y$ be a dominant, generically finite morphism of finite type of integral schemes. Show that there is an open dense subset $U \subseteq Y$ such that the induced morphism $f^{-1}(U) \rightarrow U$ is finite.[^hint.2.3.7]

[^hint.2.3.7]: 
Hint: First show that the function field of $X$ is a finite field extension of the function field of $Y$.

### II.3.8. Normalization. 

A scheme is **normal** if all of its local rings are integrally closed domains. Let $X$ be an integral scheme. For each open affine subset $U=\operatorname{Spec} A$ of $X$, let $\tilde{A}$ be the integral closure of $A$ in its quotient field, and let $\tilde{U}=\operatorname{Spec} \tilde{A}$. 

Show that one can glue the schemes $\tilde{U}$ to obtain a normal integral scheme $\tilde{X}$, called the **normalization** of $X$. 

Show also that there is a morphism $\tilde{X} \rightarrow X$, having the following universal property: for every normal integral scheme $Z$, and for every dominant morphism $f: Z \rightarrow X, f$ factors uniquely through $\tilde{X}$. If $X$ is of finite type over a field $k$, then the morphism $\tilde{X} \rightarrow X$ is a finite morphism. This generalizes (I, Ex. 3.17).

### II.3.9. The Topological Space of a Product. 

Recall that in the category of varieties, the Zariski topology on the product of two varieties is not equal to the product topology (I, Ex. 1.4). Now we see that in the category of schemes, the underlying point set of a product of schemes is not even the product set.

a. Let $k$ be a field, and let $\mathbf{A}_k^1=\operatorname{Spec} k[x]$ be the affine line over $k$. Show that $\mathbf{A}_k^1 \fiberprod{\spec k} \mathbf{A}_k^1 \cong \mathbf{A}_k^2$, and show that the underlying point set of the product is not the product of the underlying point sets of the factors (even if $k$ is algebraically closed).

b. Let $k$ be a field, let $s$ and $t$ be indeterminates over $k$. Then $\spec k(s)$, $\spec k(t)$, and  $\spec k$ are all one-point spaces. Describe the product scheme Spec $k(s) \fiberprod{\spec k} \operatorname{Spec} k(t)$.

### II.3.10. Fibres of a Morphism.

a. If $f: X \rightarrow Y$ is a morphism, and $y \in Y$ a point, show that $\operatorname{sp}\left(X_y\right)$ is homeomorphic to $f^{-1}(y)$ with the induced topology.

b. Let $X=\operatorname{Spec} k[s, t] /\left(s-t^2\right)$, let $Y=\operatorname{Spec} k[s]$, and let $f: X \rightarrow Y$ be the morphism defined by sending $s \rightarrow s$. 
	- If $y \in Y$ is the point $a \in k$ with $a \neq 0$, show that the fibre $X_y$ consists of two points, with residue field $k$. 
	- If $y \in Y$ corresponds to $0 \in k$, show that the fibre $X_y$ is a nonreduced one-point scheme. 
	- If $\eta$ is the generic point of $Y$, show that $X_\eta$ is a one-point scheme, whose residue field is an extension of degree two of the residue field of $\eta$. (Assume $k$ algebraically closed.)

### II.3.11. Closed Subschemes.

a. Closed immersions are stable under base extension: if $f: Y \rightarrow X$ is a closed immersion, and if $X^{\prime} \rightarrow X$ is any morphism, then $f^{\prime}: Y \times_X X^{\prime} \rightarrow X^{\prime}$ is also a closed immersion.

(b) \* If $Y$ is a closed subscheme of an affine scheme $X=\operatorname{Spec} A$, then $Y$ is also affine, and in fact $Y$ is the closed subscheme determined by a suitable ideal $\mathfrak{a} \subseteq A$ as the image of the closed immersion $\operatorname{Spec} A / \mathfrak{a} \rightarrow \operatorname{Spec} A$.[^hint.2.3.something.b][^4]

c. Let $Y$ be a closed subset of a scheme $X$, and give $Y$ the reduced induced subscheme structure. If $Y^{\prime}$ is any other closed subscheme of $X$ with the same underlying topological space, show that the closed immersion $Y \rightarrow X$ factors through $Y^{\prime}$. 
	We express this property by saying that **the reduced induced structure is the smallest subscheme structure on a closed subset**.

d. Let $f: Z \rightarrow X$ be a morphism. Then there is a unique closed subscheme $Y$ of $X$ with the following property: the morphism $f$ factors through $Y$, and if $Y^{\prime}$ is any other closed subscheme of $X$ through which $f$ factors, then $Y \rightarrow X$ factors through $Y^{\prime}$ also. 
	We call $Y$ the **scheme-theoretic image** of $f$. If $Z$ is a reduced scheme, then $Y$ is just the reduced induced structure on the closure of the image $f(Z)$.

[^hint.2.3.something.b]: 
Hints: 
	First show that $Y$ can be covered by a finite number of open affine subsets of the form $D\left(f_i\right) \cap Y$, with $f_i \in A$. By adding some more $f_i$ with $D\left(f_i\right) \cap Y=\varnothing$, if necessary, show that we may assume that the $D\left(f_i\right)$ cover $X$. 
	Next show that $f_1, \ldots, f_r$ generate the unit ideal of $A$. Then use (Ex. 2.17b) to show that $Y$ is affine, and (Ex. 2.18d) to show that $Y$ comes from an ideal $\mathfrak{a} \subseteq A$.] 


### II.3.12. Closed Subschemes of Proj $S$.

a. Let $\varphi: S \rightarrow T$ be a surjective homomorphism of graded rings, preserving degrees. Show that the open set $U$ of (Ex. 2.14) is equal to  $\Proj T$, and the morphism $f: \Proj T\to \Proj S$ is a closed immersion.

b. If $I \subseteq S$ is a homogeneous ideal, take $T=S / I$ and let $Y$ be the closed subscheme of $X=\operatorname{Proj} S$ defined as image of the closed immersion $\operatorname{Proj} S / I \rightarrow X$. Show that different homogeneous ideals can give rise to the same closed subscheme. 
	For example, let $d_0$ be an integer, and let $I^{\prime}=\bigoplus_{d \geqslant d_0} I_d$. Show that $I$ and $I^{\prime}$ determine the same closed subscheme.[^see_later_closed]

[^see_later_closed]: 
We will see later (5.16) that every closed subscheme of $X$ comes from a homogeneous ideal $I$ of $S$ (at least in the case where $S$ is a polynomial ring over $S_0$ ).

### II.3.13. Properties of Morphisms of Finite Type.

a. A closed immersion is a morphism of finite type.

b. A quasi-compact open immersion (Ex. 3.2) is of finite type.

c. A composition of two morphisms of finite type is of finite type.

d. Morphisms of finite type are stable under base extension.

e. If $X$ and $Y$ are schemes of finite type over $S$, then $X \times{ }_S Y$ is of finite type over $S$.

f. If $X \stackrel{f}{\rightarrow} Y \stackrel{g}{\rightarrow} Z$ are two morphisms, and if $f$ is quasi-compact, and $g \circ f$ is of finite type, then $f$ is of finite type.

g. If $f: X \rightarrow Y$ is a morphism of finite type, and if $Y$ is noetherian, then $X$ is noetherian.

### II.3.14

If $X$ is a scheme of finite type over a field, show that the closed points of $X$ are dense. Give an example to show that this is not true for arbitrary schemes.

### II.3.15

Let $X$ be a scheme of finite type over a field $k$ (not necessarily algebraically closed).

a. Show that the following three conditions are equivalent 
(in which case we say that $X$ is **geometrically irreducible**):

    - i: $X \fiberprod{k} \bar{k}$ is irreducible, where $\bar{k}$ denotes the algebraic closure of $k$.[^abuse_notation_fiberprod]
    - ii: $X \fiberprod{k} k_s$ is irreducible, where $k_s$ denotes the separable closure of $k$.
    - iii: $X \fiberprod{k} K$ is irreducible for every extension field $K$ of $k$.

b. Show that the following three conditions are equivalent 
(in which case we say $X$ is **geometrically reduced**):

    - i: $X \fiberprod{k} \bar{k}$ is reduced.
    - ii: $X \fiberprod{k} k_p$ is reduced, where $k_p$ denotes the perfect closure of $k$.
    - iii: $X \fiberprod{k} K$ is reduced for all extension fields $K$ of $k$.

c. We say that $X$ is **geometrically integral** if $X \fiberprod{k} \bar{k}$ is integral. 
	Give examples of integral schemes which are neither geometrically irreducible nor geometrically reduced.

[^abuse_notation_fiberprod]: 
By abuse of notation, we write $X \fiberprod{k} \bar{k}$ to denote $X \fiberprod{\spec k} \spec \bar{k}$.

### II.3.16. Noetherian Induction. 

Let $X$ be a noetherian topological space, and let $\mathscr{P}$ be a property of closed subsets of $X$. Assume that for any closed subset $Y$ of $X$, if $\mathscr{P}$ holds for every proper closed subset of $Y$, then $\mathscr{P}$ holds for $Y$. (In particular, $\mathscr{P}$ must hold for the empty set.) Then $\mathscr{P}$ holds for $X$.


### II.3.17. Zariski Spaces. 

A topological space $X$ is a **Zariski space** if it is noetherian and every (nonempty) closed irreducible subset has a unique generic point (Ex. 2.9).

For example, let $R$ be a discrete valuation ring, and let $T=\operatorname{sp}(\operatorname{Spec} R)$. Then $T$ consists of two points $t_0=$ the maximal ideal, $t_1=$ the zero ideal. The open subsets are $\varnothing,\left\{t_1\right\}$, and $T$. This is an irreducible Zariski space with generic point $t_1$.

a. Show that if $X$ is a noetherian scheme, then $\operatorname{sp}(X)$ is a Zariski space.

b. Show that any minimal nonempty closed subset of a Zariski space consists of one point. We call these closed points.

c. Show that a Zariski space $X$ satisfies the axiom $T_0$ :given any two distinct points of $X$, there is an open set containing one but not the other.

d. If $X$ is an irreducible Zariski space, then its generic point is contained in every nonempty open subset of $X$ of $x_1$, or that $x_1$ is a generization of $x_0$. Now let $X$ be a Zariski space. 

      - Show that the minimal points, for the partial ordering determined by $x_1>x_0$ if $x_1 \leftrightarrow$ $x_0$, are the closed points, and the maximal points are the generic points of the irreducible components of $X$. 

      - Show also that a closed subset contains every specialization of any of its points. (We say closed subsets are **stable under specialization**.) Similarly, open subsets are stable under generization.

f. Let $t$ be the functor on topological spaces introduced in the proof of (2.6). If $X$ is a noetherian topological space, show that $t(X)$ is a Zariski space. Furthermore $X$ itself is a Zariski space if and only if the map $\alpha: X \rightarrow t(X)$ is a homeomorphism.

### II.3.18 Constructible Sets. 

Let $X$ be a Zariski topological space. A constructible subset of $X$ is a subset which belongs to the smallest family $\mathscr{F}$ of subsets such that (1) every open subset is in $\mathscr{F},(2)$ a finite intersection of elements of $\mathscr{F}$ is in $\mathscr{F}$, and (3) the complement of an element of $\mathfrak{F}$ is in $\mathfrak{F}$.

a. A subset of $X$ is locally closed if it is the intersection of an open subset with a closed subset. Show that a subset of $X$ is constructible if and only if it can be written as a finite disjoint union of locally closed subsets.

b. Show that a constructible subset of an irreducible Zariski space $X$ is dense if and only if it contains the generic point. Furthermore, in that case it contains a nonempty open subset.

c. A subset $S$ of $X$ is closed if and only if it is constructible and stable under specialization. Similarly, a subset $T$ of $X$ is open if and only if it is constructible and stable under generization.

d. If $f: X \rightarrow Y$ is a continuous map of Zariski spaces, then the inverse image of any constructible subset of $Y$ is a constructible subset of $X$.

### II.3.19

Let $f: X \rightarrow Y$ be a morphism of finite type of noetherian schemes. Then the image of any constructible subset of $X$ is a constructible subset of $Y$. In particular, $f(X)$, which need not be either open or closed, is a constructible subset of $Y$.[^real_importance]

Prove this theorem in the following steps.

a. Reduce to showing that $f(X)$ itself is constructible, in the case where $X$ and $Y$ are affine, integral noetherian schemes, and $f$ is a dominant morphism.

b. \* In that case, show that $f(X)$ contains a nonempty open subset of $Y$ by using the following result from commutative algebra: let $A \subseteq B$ be an inclusion of noetherian integral domains, such that $B$ is a finitely generated $A$-algebra. 
	Then given a nonzero element $b \in B$, there is a nonzero element $a \in A$ with the following property: if $\varphi: A \rightarrow K$ is any homomorphism of $A$ to an algebraically closed field $K$, such that $\varphi(a) \neq 0$, then $\varphi$ extends to a homomorphism $\varphi^{\prime}$ of $B$ into $K$, such that $\varphi^{\prime}(b) \neq 0$.[^hint.2.3.19.b]

c. Now use noetherian induction on $Y$ to complete the proof.

d. Give some examples of morphisms $f: X \rightarrow Y$ of varieties over an algebraically closed field $k$, to show that $f(X)$ need not be either open or closed.

[^hint.2.3.19.b]: 
Hint: Prove this algebraic result by induction on the number of generators of $B$ over $A$. For the case of one generator, prove the result directly. In the application, take $b=1$.

[^real_importance]: 
The real importance of the notion of constructible subsets derives from the following theorem of Chevalley-see Cartan and Chevalley (1, exposé 7) and see also Matsumura (2, Ch. 2, 6). 

### II.3.20. Dimension. 

Let $X$ be an integral scheme of finite type over a field $k$ (not necessarily algebraically closed). Use appropriate results from I.1 to prove the following.

a. For any closed point $P \in X, \operatorname{dim} X=\operatorname{dim} \mathcal{O}_P$, where for rings, we always mean the Krull dimension.

b. Let $K(X)$ be the function field of $X$ (Ex. 3.6). Then $$\operatorname{dim} X=\trdeg K(X)/k.$$

c. If $Y$ is a closed subset of $X$, then 
$$\operatorname{codim}(Y, X)=\inf \left\{\operatorname{dim} \mathcal{O}_{P, X} \mid P  \in Y\right\}.$$

d. If $Y$ is a closed subset of $X$, then 
$$\operatorname{dim} Y+\operatorname{codim}(Y, X)=\operatorname{dim} X.$$

e. If $U$ is a nonempty open subset of $X$, then $\operatorname{dim} U=\operatorname{dim} X$.

f. If $k \subseteq k^{\prime}$ is a field extension, then every irreducible component of $X^{\prime}=X \fiberprod{k} k^{\prime}$ has dimension $=\operatorname{dim} X$.

### II.3.21

Let $R$ be a discrete valuation ring containing its residue field $k$. Let $X=$ Spec $R[t]$ be the affine line over Spec $R$. Show that statements (a), (d), (e) of (Ex. 3.20) are false for $X$.

### 3.22. \* Dimension of the Fibres of a Morphism. 

Let $f: X \rightarrow Y$ be a dominant morphism of integral schemes of finite type over a field $k$.

a. Let $Y^{\prime}$ be a closed irreducible subset of $Y$, whose generic point $\eta^{\prime}$ is contained in $f(X)$. Let $Z$ be any irreducible component of $f^{-1}\left(Y^{\prime}\right)$, such that $\eta^{\prime} \in f(Z)$, and show that $\operatorname{codim}(Z, X) \leqslant \operatorname{codim}\left(Y^{\prime}, Y\right)$.

b. Let $e=\operatorname{dim} X-\operatorname{dim} Y$ be the relative dimension of $X$ over $Y$. For any point $y \in f(X)$, show that every irreducible component of the fibre $X_y$ has dimension $\geqslant e$.[^hint2.3.qqqq.b]

c. Show that there is a dense open subset $U \subseteq X$, such that for any $y \in f(U)$, $\operatorname{dim} U_y=e$.[^hint_2.3.smth.c]
  
d. Going back to our original morphism $f: X \rightarrow Y$, for any integer $h$, let $E_h$ be the set of points $x \in X$ such that, letting $y=f(x)$, there is an irreducible component $Z$ of the fibre $X_y$, containing $x$, and having $\operatorname{dim} Z \geqslant h$. Show that 
	- 1) $E_e=X$[^1]; 
	- 2) if $h>e$, then $E_h$ is not dense in $X$[^2]; and 
	- 3) $E_h$ is closed, for all $h$[^3].

e. Prove the following theorem of Chevalley-see Cartan and Chevalley (1, exposé 8): For each integer $h$, let $C_h$ be the set of points $y \in Y$ such that dim $X_y=h$. Then the subsets $C_h$ are constructible, and $C_e$ contains an open dense subset of $Y$.

[^hint2.3.qqqq.b]: 
Hint: Let $Y^{\prime}=\{y\}^{-}$, and use (a) and (Ex. 3.20b).

[^hint_2.3.smth.c]: 
Hint: First reduce to the case where $X$ and $Y$ are affine, say $X=\operatorname{Spec} A$ and $Y=\operatorname{Spec} B$. Then $A$ is a finitely generated $B$-algebra. Take $t_1, \ldots, t_e \in A$ which form a transcendence base of $K(X)$ over $K(Y)$, and let $X_1=\operatorname{Spec} B\left[t_1, \ldots, t_e\right]$. Then $X_1$ is isomorphic to affine $e$-space over $Y$, and the morphism $X \rightarrow X_1$ is generically finite. Now use (Ex. 3.7) above.

### II.3.23

If $V, W$ are two varieties over an algebraically closed field $k$, and if $V \times W$ is their product, as defined in (I, Ex. 3.15, 3.16), and if $t$ is the functor of $(2.6)$, then $t(V \times W)=t(V) \fiberprod{\spec k} t(W)$.

[^1]: Use (b) above.
[^2]: Use (c) above.
[^3]: Use induction on $\operatorname{dim} X$.
[^4]: Note: We will give another proof of this result using sheaves of ideals later (V.10).