## 29.49 Generically finite morphisms

In this section we characterize maps between schemes which are locally of finite type and which are “generically finite” in some sense.

Lemma 29.49.1. Let $X$, $Y$ be schemes. Let $f : X \to Y$ be locally of finite type. Let $\eta \in Y$ be a generic point of an irreducible component of $Y$. The following are equivalent:

the set $f^{-1}(\{ \eta \} )$ is finite,

there exist affine opens $U_ i \subset X$, $i = 1, \ldots , n$ and $V \subset Y$ with $f(U_ i) \subset V$, $\eta \in V$ and $f^{-1}(\{ \eta \} ) \subset \bigcup U_ i$ such that each $f|_{U_ i} : U_ i \to V$ is finite.

If $f$ is quasi-separated, then these are also equivalent to

there exist affine opens $V \subset Y$, and $U \subset X$ with $f(U) \subset V$, $\eta \in V$ and $f^{-1}(\{ \eta \} ) \subset U$ such that $f|_ U : U \to V$ is finite.

If $f$ is quasi-compact and quasi-separated, then these are also equivalent to

there exists an affine open $V \subset Y$, $\eta \in V$ such that $f^{-1}(V) \to V$ is finite.

**Proof.**
The question is local on the base. Hence we may replace $Y$ by an affine neighbourhood of $\eta $, and we may and do assume throughout the proof below that $Y$ is affine, say $Y = \mathop{\mathrm{Spec}}(R)$.

It is clear that (2) implies (1). Assume that $f^{-1}(\{ \eta \} ) = \{ \xi _1, \ldots , \xi _ n\} $ is finite. Choose affine opens $U_ i \subset X$ with $\xi _ i \in U_ i$. By Algebra, Lemma 10.121.10 we see that after replacing $Y$ by a standard open in $Y$ each of the morphisms $U_ i \to Y$ is finite. In other words (2) holds.

It is clear that (3) implies (1). Assume $f^{-1}(\{ \eta \} ) = \{ \xi _1, \ldots , \xi _ n\} $ and assume that $f$ is quasi-separated. Since $Y$ is affine this implies that $X$ is quasi-separated. Since each $\xi _ i$ maps to a generic point of an irreducible component of $Y$, we see that each $\xi _ i$ is a generic point of an irreducible component of $X$. By Properties, Lemma 28.29.1 we can find an affine open $U \subset X$ containing each $\xi _ i$. By Algebra, Lemma 10.121.10 we see that after replacing $Y$ by a standard open in $Y$ the morphisms $U \to Y$ is finite. In other words (3) holds.

It is clear that (4) implies all of (1) – (3) with no further assumptions on $f$. Suppose that $f$ is quasi-compact and quasi-separated. We have to show that the equivalent conditions (1) – (3) imply (4). Let $U$, $V$ be as in (3). Replace $Y$ by $V$. Since $f$ is quasi-compact and $Y$ is quasi-compact (being affine) we see that $X$ is quasi-compact. Hence $Z = X \setminus U$ is quasi-compact, hence the morphism $f|_ Z : Z \to Y$ is quasi-compact. By construction of $Z$ we see that $\eta \not\in f(Z)$. Hence by Lemma 29.8.4 we see that there exists an affine open neighbourhood $V'$ of $\eta $ in $Y$ such that $f^{-1}(V') \cap Z = \emptyset $. Then we have $f^{-1}(V') \subset U$ and this means that $f^{-1}(V') \to V'$ is finite.
$\square$

Example 29.49.2. Let $A = \prod _{n \in \mathbf{N}} \mathbf{F}_2$. Every element of $A$ is an idempotent. Hence every prime ideal is maximal with residue field $\mathbf{F}_2$. Thus the topology on $X = \mathop{\mathrm{Spec}}(A)$ is totally disconnected and quasi-compact. The projection maps $A \to \mathbf{F}_2$ define open points of $\mathop{\mathrm{Spec}}(A)$. It cannot be the case that all the points of $X$ are open since $X$ is quasi-compact. Let $x \in X$ be a closed point which is not open. Then we can form a scheme $Y$ which is two copies of $X$ glued along $X \setminus \{ x\} $. In other words, this is $X$ with $x$ doubled, compare Schemes, Example 26.14.3. The morphism $f : Y \to X$ is quasi-compact, finite type and has finite fibres but is not quasi-separated. The point $x \in X$ is a generic point of an irreducible component of $X$ (since $X$ is totally disconnected). But properties (3) and (4) of Lemma 29.49.1 do not hold. The reason is that for any open neighbourhood $x \in U \subset X$ the inverse image $f^{-1}(U)$ is not affine because functions on $f^{-1}(U)$ cannot separate the two points lying over $x$ (proof omitted; this is a nice exercise). Hence the condition that $f$ is quasi-separated is necessary in parts (3) and (4) of the lemma.

Lemma 29.49.4. Let $X$, $Y$ be schemes. Let $f : X \to Y$ be locally of finite type. Let $X^0$, resp. $Y^0$ denote the set of generic points of irreducible components of $X$, resp. $Y$. Let $\eta \in Y^0$. The following are equivalent

$f^{-1}(\{ \eta \} ) \subset X^0$,

$f$ is quasi-finite at all points lying over $\eta $,

$f$ is quasi-finite at all $\xi \in X^0$ lying over $\eta $.

**Proof.**
Condition (1) implies there are no specializations among the points of the fibre $X_\eta $. Hence (2) holds by Lemma 29.19.6. The implication (2) $\Rightarrow $ (3) is immediate. Since $\eta $ is a generic point of $Y$, the generic points of $X_\eta $ are generic points of $X$. Hence (3) and Lemma 29.19.6 imply the generic points of $X_\eta $ are also closed. Thus all points of $X_\eta $ are generic and we see that (1) holds.
$\square$

Lemma 29.49.5. Let $X$, $Y$ be schemes. Let $f : X \to Y$ be locally of finite type. Let $X^0$, resp. $Y^0$ denote the set of generic points of irreducible components of $X$, resp. $Y$. Assume

$X^0$ and $Y^0$ are finite and $f^{-1}(Y^0) = X^0$,

either $f$ is quasi-compact or $f$ is separated.

Then there exists a dense open $V \subset Y$ such that $f^{-1}(V) \to V$ is finite.

**Proof.**
Since $Y$ has finitely many irreducible components, we can find a dense open which is a disjoint union of its irreducible components. Thus we may assume $Y$ is irreducible affine with generic point $\eta $. Then the fibre over $\eta $ is finite as $X^0$ is finite.

Assume $f$ is separated and $Y$ irreducible affine. Choose $V \subset Y$ and $U \subset X$ as in Lemma 29.49.1 part (3). Since $f|_ U : U \to V$ is finite, we see that $U \subset f^{-1}(V)$ is closed as well as open (Lemmas 29.39.7 and 29.42.11). Thus $f^{-1}(V) = U \amalg W$ for some open subscheme $W$ of $X$. However, since $U$ contains all the generic points of $X$ we conclude that $W = \emptyset $ as desired.

Assume $f$ is quasi-compact and $Y$ irreducible affine. Then $X$ is quasi-compact, hence there exists a dense open subscheme $U \subset X$ which is separated (Properties, Lemma 28.29.3). Since the set of generic points $X^0$ is finite, we see that $X^0 \subset U$. Thus $\eta \not\in f(X \setminus U)$. Since $X \setminus U \to Y$ is quasi-compact, we conclude that there is a nonempty open $V \subset Y$ such that $f^{-1}(V) \subset U$, see Lemma 29.8.3. After replacing $X$ by $f^{-1}(V)$ and $Y$ by $V$ we reduce to the separated case which we dealt with in the preceding paragraph.
$\square$

Lemma 29.49.6. Let $X$, $Y$ be schemes. Let $f : X \to Y$ be a birational morphism between schemes which have finitely many irreducible components. Assume

either $f$ is quasi-compact or $f$ is separated, and

either $f$ is locally of finite type and $Y$ is reduced or $f$ is locally of finite presentation.

Then there exists a dense open $V \subset Y$ such that $f^{-1}(V) \to V$ is an isomorphism.

**Proof.**
By Lemma 29.49.5 we may assume that $f$ is finite. Since $Y$ has finitely many irreducible components, we can find a dense open which is a disjoint union of its irreducible components. Thus we may assume $Y$ is irreducible. By Lemma 29.48.5 we find a nonempty open $U \subset X$ such that $f|_ U : U \to Y$ is an open immersion. After removing the closed (as $f$ finite) subset $f(X \setminus U)$ from $Y$ we see that $f$ is an isomorphism.
$\square$

Lemma 29.49.7. Let $X$, $Y$ be integral schemes. Let $f : X \to Y$ be locally of finite type. Assume $f$ is dominant. The following are equivalent:

the extension $R(Y) \subset R(X)$ has transcendence degree $0$,

the extension $R(Y) \subset R(X)$ is finite,

there exist nonempty affine opens $U \subset X$ and $V \subset Y$ such that $f(U) \subset V$ and $f|_ U : U \to V$ is finite, and

the generic point of $X$ is the only point of $X$ mapping to the generic point of $Y$.

If $f$ is separated or if $f$ is quasi-compact, then these are also equivalent to

there exists a nonempty affine open $V \subset Y$ such that $f^{-1}(V) \to V$ is finite.

**Proof.**
Choose any affine opens $\mathop{\mathrm{Spec}}(A) = U \subset X$ and $\mathop{\mathrm{Spec}}(R) = V \subset Y$ such that $f(U) \subset V$. Then $R$ and $A$ are domains by definition. The ring map $R \to A$ is of finite type (Lemma 29.14.2). By Lemma 29.8.5 the generic point of $X$ maps to the generic point of $Y$ hence $R \to A$ is injective. Let $K = R(Y)$ be the fraction field of $R$ and $L = R(X)$ the fraction field of $A$. Then $K \subset L$ is a finitely generated field extension. Hence we see that (1) is equivalent to (2).

Suppose (2) holds. Let $x_1, \ldots , x_ n \in A$ be generators of $A$ over $R$. By assumption there exist nonzero polynomials $P_ i(X) \in R[X]$ such that $P_ i(x_ i) = 0$. Let $f_ i \in R$ be the leading coefficient of $P_ i$. Then we conclude that $R_{f_1 \ldots f_ n} \to A_{f_1 \ldots f_ n}$ is finite, i.e., (3) holds. Note that (3) implies (2). So now we see that (1), (2) and (3) are all equivalent.

Let $\eta $ be the generic point of $X$, and let $\eta ' \in Y$ be the generic point of $Y$. Assume (4). Then $\dim _\eta (X_{\eta '}) = 0$ and we see that $R(X) = \kappa (\eta )$ has transcendence degree $0$ over $R(Y) = \kappa (\eta ')$ by Lemma 29.27.1. In other words (1) holds. Assume the equivalent conditions (1), (2) and (3). Suppose that $x \in X$ is a point mapping to $\eta '$. As $x$ is a specialization of $\eta $, this gives inclusions $R(Y) \subset \mathcal{O}_{X, x} \subset R(X)$, which implies $\mathcal{O}_{X, x}$ is a field, see Algebra, Lemma 10.35.19. Hence $x = \eta $. Thus we see that (1) – (4) are all equivalent.

It is clear that (5) implies (3) with no additional assumptions on $f$. What remains is to prove that if $f$ is either separated or quasi-compact, then the equivalent conditions (1) – (4) imply (5). This follows from Lemma 29.49.5.
$\square$

Definition 29.49.8. Let $X$ and $Y$ be integral schemes. Let $f : X \to Y$ be locally of finite type and dominant. Assume $[R(X) : R(Y)] < \infty $, or any other of the equivalent conditions (1) – (4) of Lemma 29.49.7. Then the positive integer

\[ \text{deg}(X/Y) = [R(X) : R(Y)] \]

is called the *degree of $X$ over $Y$*.

It is possible to extend this notion to a morphism $f : X \to Y$ if (a) $Y$ is integral with generic point $\eta $, (b) $f$ is locally of finite type, and (c) $f^{-1}(\{ \eta \} )$ is finite. In this case we can define

\[ \deg (X/Y) = \sum \nolimits _{\xi \in X, \ f(\xi ) = \eta } \dim _{R(Y)} (\mathcal{O}_{X, \xi }). \]

Namely, given that $R(Y) = \kappa (\eta ) = \mathcal{O}_{Y, \eta }$ (Lemma 29.47.5) the dimensions above are finite by Lemma 29.49.1 above. However, for most applications the definition given above is the right one.

Lemma 29.49.9. Let $X$, $Y$, $Z$ be integral schemes. Let $f : X \to Y$ and $g : Y \to Z$ be dominant morphisms locally of finite type. Assume that $[R(X) : R(Y)] < \infty $ and $[R(Y) : R(Z)] < \infty $. Then

\[ \deg (X/Z) = \deg (X/Y) \deg (Y/Z). \]

**Proof.**
This comes from the multiplicativity of degrees in towers of finite extensions of fields, see Fields, Lemma 9.7.7.
$\square$

Definition 29.49.11. Let $X$ be an integral scheme. A *modification of $X$* is a birational proper morphism $f : X' \to X$ with $X'$ integral.

Let $f : X' \to X$ be a modification as in the definition. By Lemma 29.49.7 there exists a nonempty $U \subset X$ such that $f^{-1}(U) \to U$ is finite. By generic flatness (Proposition 29.26.1) we may assume $f^{-1}(U) \to U$ is flat and of finite presentation. So $f^{-1}(U) \to U$ is finite locally free (Lemma 29.46.2). Since $f$ is birational, the degree of $X'$ over $X$ is $1$. Hence $f^{-1}(U) \to U$ is finite locally free of degree $1$, in other words it is an isomorphism. Thus we can *redefine* a modification to be a proper morphism $f : X' \to X$ of integral schemes such that $f^{-1}(U) \to U$ is an isomorphism for some nonempty open $U \subset X$.

reference
Definition 29.49.12. Let $X$ be an integral scheme. An *alteration of $X$* is a proper dominant morphism $f : Y \to X$ with $Y$ integral such that $f^{-1}(U) \to U$ is finite for some nonempty open $U \subset X$.

This is the definition as given in [alterations], except that here we do not require $X$ and $Y$ to be Noetherian. Arguing as above we see that an alteration is a proper dominant morphism $f : Y \to X$ of integral schemes which induces a finite extension of function fields, i.e., such that the equivalent conditions of Lemma 29.49.7 hold.

## Comments (0)