## 29.51 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 29.51.1. 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:

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

2. 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

1. 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

1. 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 10.122.10 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 $f$ is quasi-separated and (1). Write $f^{-1}(\{ \eta \} ) = \{ \xi _1, \ldots , \xi _ n\}$. There are no specializations among the $\xi _ i$ by Lemma 29.20.7. Since each $\xi _ i$ maps to the generic point $\eta$ of an irreducible component of $Y$, there cannot be a nontrivial specialization $\xi \leadsto \xi _ i$ in $X$ (since $\xi$ would map to $\eta$ as well). We conclude each $\xi _ i$ is a generic point of an irreducible component of $X$. Since $Y$ is affine and $f$ quasi-separated we see $X$ is quasi-separated. By Properties, Lemma 28.29.1 we can find an affine open $U \subset X$ containing each $\xi _ i$. By Algebra, Lemma 10.122.10 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 29.51.2. 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 29.51.1 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.

Remark 29.51.3. An alternative to Lemma 29.51.1 is the statement that a quasi-finite morphism is finite over a dense open of the target. This will be shown in More on Morphisms, Lemma 37.41.2.

Lemma 29.51.4. 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

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

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

3. $f$ 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 29.20.6. 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 29.20.6 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 29.51.5. 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

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

2. 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 29.51.1 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 29.41.7 and 29.44.11). 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 29.51.6. Let $X$, $Y$ be schemes. Let $f : X \to Y$ be a birational morphism between schemes which have finitely many irreducible components. Assume

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

2. 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 29.51.5 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 29.50.5 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 29.51.7. Let $X$, $Y$ be integral schemes. Let $f : X \to Y$ be locally of finite type. Assume $f$ is dominant. The following are equivalent:

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

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

3. 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

4. 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

1. 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 29.15.2). 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 29.28.1. 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 10.36.19. 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 29.51.5. $\square$

Definition 29.51.8. 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 29.51.7. 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 29.49.5) the dimensions above are finite by Lemma 29.51.1 above. However, for most applications the definition given above is the right one.

Lemma 29.51.9. 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. $\square$

Remark 29.51.10. Let $f : X \to Y$ be a morphism of schemes which is locally of finite type. There are (at least) two properties that we could use to define generically finite morphisms. These correspond to whether you want the property to be local on the source or local on the target:

1. (Local on the target; suggested by Ravi Vakil.) Assume every quasi-compact open of $Y$ has finitely many irreducible components (for example if $Y$ is locally Noetherian). The requirement is that the inverse image of each generic point is finite, see Lemma 29.51.1.

2. (Local on the source.) The requirement is that there exists a dense open $U \subset X$ such that $U \to Y$ is locally quasi-finite.

In case (1) the requirement can be formulated without the auxiliary condition on $Y$, but probably doesn't give the right notion for general schemes. Property (2) as formulated doesn't imply that the fibres over generic points are finite; however, if $f$ is quasi-compact and $Y$ is as in (1) then it does.

Definition 29.51.11. 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 29.51.7 there exists a nonempty $U \subset X$ such that $f^{-1}(U) \to U$ is finite. By generic flatness (Proposition 29.27.1) 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 29.48.2). 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 $U \subset X$.

Definition 29.51.12. 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 , 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 29.51.7 hold.

Comment #5084 by GGTTTTLL on

In the proof of lemma 02NW, third paragraph, it says "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 $X$. " does not seem correct in general, take any morphism to $Spec k$ for example. Maybe say: Since by (2), $\xi_i\in U_i$ maps to $\eta$ by finite morphisms, $\xi_i$ is a generic point of $X$.

Comment #5295 by on

What happened here probably happens all over the place: I used that if you have a scheme $S$ (locally) of finite type over a field with finitely many points, then those points are all generic points. Moreover, if this happens for a generic fibre $S = X_\eta$, then those points are all generic points of irreducible components of $X$. This is just part of my arsenal of things that are true and it is almost impossible to see (for me) that I am using something like this. So thanks to GGTTTTLL for catching this. I fixed it in a slightly different way, see this commit.

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