## 29.42 Integral and finite morphisms

Recall that a ring map $R \to A$ is said to be *integral* if every element of $A$ satisfies a monic equation with coefficients in $R$. Recall that a ring map $R \to A$ is said to be *finite* if $A$ is finite as an $R$-module. See Algebra, Definition 10.35.1.

Definition 29.42.1. Let $f : X \to S$ be a morphism of schemes.

We say that $f$ is *integral* if $f$ is affine and if for every affine open $\mathop{\mathrm{Spec}}(R) = V \subset S$ with inverse image $\mathop{\mathrm{Spec}}(A) = f^{-1}(V) \subset X$ the associated ring map $R \to A$ is integral.

We say that $f$ is *finite* if $f$ is affine and if for every affine open $\mathop{\mathrm{Spec}}(R) = V \subset S$ with inverse image $\mathop{\mathrm{Spec}}(A) = f^{-1}(V) \subset X$ the associated ring map $R \to A$ is finite.

It is clear that integral/finite morphisms are separated and quasi-compact. It is also clear that a finite morphism is a morphism of finite type. Most of the lemmas in this section are completely standard. But note the fun Lemma 29.42.7 at the end of the section.

Lemma 29.42.2. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

The morphism $f$ is integral.

There exists an affine open covering $S = \bigcup U_ i$ such that each $f^{-1}(U_ i)$ is affine and $\mathcal{O}_ S(U_ i) \to \mathcal{O}_ X(f^{-1}(U_ i))$ is integral.

There exists an open covering $S = \bigcup U_ i$ such that each $f^{-1}(U_ i) \to U_ i$ is integral.

Moreover, if $f$ is integral then for every open subscheme $U \subset S$ the morphism $f : f^{-1}(U) \to U$ is integral.

**Proof.**
See Algebra, Lemma 10.35.14. Some details omitted.
$\square$

Lemma 29.42.3. Let $f : X \to S$ be a morphism of schemes. The following are equivalent:

The morphism $f$ is finite.

There exists an affine open covering $S = \bigcup U_ i$ such that each $f^{-1}(U_ i)$ is affine and $\mathcal{O}_ S(U_ i) \to \mathcal{O}_ X(f^{-1}(U_ i))$ is finite.

There exists an open covering $S = \bigcup U_ i$ such that each $f^{-1}(U_ i) \to U_ i$ is finite.

Moreover, if $f$ is finite then for every open subscheme $U \subset S$ the morphism $f : f^{-1}(U) \to U$ is finite.

**Proof.**
See Algebra, Lemma 10.35.14. Some details omitted.
$\square$

Lemma 29.42.4. A finite morphism is integral. An integral morphism which is locally of finite type is finite.

**Proof.**
See Algebra, Lemma 10.35.3 and Lemma 10.35.5.
$\square$

Lemma 29.42.5. A composition of finite morphisms is finite. Same is true for integral morphisms.

**Proof.**
See Algebra, Lemmas 10.7.3 and 10.35.6.
$\square$

Lemma 29.42.6. A base change of a finite morphism is finite. Same is true for integral morphisms.

**Proof.**
See Algebra, Lemma 10.35.13.
$\square$

slogan
Lemma 29.42.7. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

$f$ is integral, and

$f$ is affine and universally closed.

**Proof.**
Assume (1). An integral morphism is affine by definition. A base change of an integral morphism is integral so in order to prove (2) it suffices to show that an integral morphism is closed. This follows from Algebra, Lemmas 10.35.22 and 10.40.6.

Assume (2). We may assume $f$ is the morphism $f : \mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(R)$ coming from a ring map $R \to A$. Let $a$ be an element of $A$. We have to show that $a$ is integral over $R$, i.e. that in the kernel $I$ of the map $R[x] \to A$ sending $x$ to $a$ there is a monic polynomial. Consider the ring $B = A[x]/(ax -1)$ and let $J$ be the kernel of the composition $R[x]\to A[x] \to B$. If $f\in J$ there exists $q\in A[x]$ such that $f = (ax-1)q$ in $A[x]$ so if $f = \sum _ i f_ ix^ i$ and $q = \sum _ iq_ ix^ i$, for all $i \geq 0$ we have $f_ i = aq_{i-1} - q_ i$. For $n \geq \deg q + 1$ the polynomial

\[ \sum \nolimits _{i \geq 0} f_ i x^{n - i} = \sum \nolimits _{i \geq 0} (a q_{i - 1} - q_ i) x^{n - i} = (a - x) \sum \nolimits _{i \geq 0} q_ i x^{n - i - 1} \]

is clearly in $I$; if $f_0 = 1$ this polynomial is also monic, so we are reduced to prove that $J$ contains a polynomial with constant term $1$. We do it by proving $\mathop{\mathrm{Spec}}(R[x]/(J + (x))$ is empty.

Since $f$ is universally closed the base change $\mathop{\mathrm{Spec}}(A[x]) \to \mathop{\mathrm{Spec}}(R[x])$ is closed. Hence the image of the closed subset $\mathop{\mathrm{Spec}}(B) \subset \mathop{\mathrm{Spec}}(A[x])$ is the closed subset $\mathop{\mathrm{Spec}}(R[x]/J) \subset \mathop{\mathrm{Spec}}(R[x])$, see Example 29.6.4 and Lemma 29.6.3. In particular $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(R[x]/J)$ is surjective. Consider the following diagram where every square is a pullback:

\[ \xymatrix{ \mathop{\mathrm{Spec}}(B) \ar@{->>}[r]^ g & \mathop{\mathrm{Spec}}(R[x]/J) \ar[r] & \mathop{\mathrm{Spec}}(R[x])\\ \emptyset \ar[u] \ar[r] & \mathop{\mathrm{Spec}}(R[x]/(J + (x)))\ar[u] \ar[r] & \mathop{\mathrm{Spec}}(R) \ar[u]^0 } \]

The bottom left corner is empty because it is the spectrum of $R\otimes _{R[x]} B$ where the map $R[x]\to B$ sends $x$ to an invertible element and $R[x]\to R$ sends $x$ to $0$. Since $g$ is surjective this implies $\mathop{\mathrm{Spec}}(R[x]/(J + (x)))$ is empty, as we wanted to show.
$\square$

Lemma 29.42.8. Let $f : X \to S$ be an integral morphism. Then every point of $X$ is closed in its fibre.

**Proof.**
See Algebra, Lemma 10.35.20.
$\square$

Lemma 29.42.9. Let $f : X \to Y$ be an integral morphism. Then $\dim (X) \leq \dim (Y)$. If $f$ is surjective then $\dim (X) = \dim (Y)$.

**Proof.**
Since the dimension of $X$ and $Y$ is the supremum of the dimensions of the members of an affine open covering, we may assume $Y$ and $X$ are affine. The inequality follows from Algebra, Lemma 10.111.3. The equality then follows from Algebra, Lemmas 10.111.1 and 10.35.22.
$\square$

Lemma 29.42.10. A finite morphism is quasi-finite.

**Proof.**
This is implied by Algebra, Lemma 10.121.4 and Lemma 29.19.9. Alternatively, all points in fibres are closed points by Lemma 29.42.8 (and the fact that a finite morphism is integral) and use Lemma 29.19.6 (3) to see that $f$ is quasi-finite at $x$ for all $x \in X$.
$\square$

Lemma 29.42.11. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

$f$ is finite, and

$f$ is affine and proper.

**Proof.**
This follows formally from Lemma 29.42.7, the fact that a finite morphism is integral and separated, the fact that a proper morphism is the same thing as a finite type, separated, universally closed morphism, and the fact that an integral morphism of finite type is finite (Lemma 29.42.4).
$\square$

Lemma 29.42.12. A closed immersion is finite (and a fortiori integral).

**Proof.**
True because a closed immersion is affine (Lemma 29.11.9) and a surjective ring map is finite and integral.
$\square$

Lemma 29.42.13. Let $X_ i \to Y$, $i = 1, \ldots , n$ be finite morphisms of schemes. Then $X_1 \amalg \ldots \amalg X_ n \to Y$ is finite too.

**Proof.**
Follows from the algebra fact that if $R \to A_ i$, $i = 1, \ldots , n$ are finite ring maps, then $R \to A_1 \times \ldots \times A_ n$ is finite too.
$\square$

Lemma 29.42.14. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms.

If $g \circ f$ is finite and $g$ separated then $f$ is finite.

If $g \circ f$ is integral and $g$ separated then $f$ is integral.

**Proof.**
Assume $g \circ f$ is finite (resp. integral) and $g$ separated. The base change $X \times _ Z Y \to Y$ is finite (resp. integral) by Lemma 29.42.6. The morphism $X \to X \times _ Z Y$ is a closed immersion as $Y \to Z$ is separated, see Schemes, Lemma 26.21.11. A closed immersion is finite (resp. integral), see Lemma 29.42.12. The composition of finite (resp. integral) morphisms is finite (resp. integral), see Lemma 29.42.5. Thus we win.
$\square$

Lemma 29.42.15. Let $f : X \to Y$ be a morphism of schemes. If $f$ is finite and a monomorphism, then $f$ is a closed immersion.

**Proof.**
This reduces to Algebra, Lemma 10.106.6.
$\square$

Lemma 29.42.16. A finite morphism is projective.

**Proof.**
Let $f : X \to S$ be a finite morphism. Then $f_*\mathcal{O}_ X$ is a quasi-coherent $\mathcal{O}_ S$-module (Lemma 29.11.5) of finite type (by our definition of finite morphisms and Properties, Lemma 28.16.1). We claim there is a closed immersion

\[ \sigma : X \longrightarrow \mathbf{P}(f_*\mathcal{O}_ X) = \underline{\text{Proj}}_ S(\text{Sym}^*_{\mathcal{O}_ S}(f_*\mathcal{O}_ X)) \]

over $S$, which finishes the proof. Namely, we let $\sigma $ be the morphism which corresponds (via Constructions, Lemma 27.16.11) to the surjection

\[ f^*f_*\mathcal{O}_ X \longrightarrow \mathcal{O}_ X \]

coming from the adjunction map $f^*f_* \to \text{id}$. Then $\sigma $ is a closed immersion by Schemes, Lemma 26.21.10 and Constructions, Lemma 27.21.4.
$\square$

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