Definition 29.11.1. A morphism of schemes $f : X \to S$ is called *affine* if the inverse image of every affine open of $S$ is an affine open of $X$.

## 29.11 Affine morphisms

Lemma 29.11.2. An affine morphism is separated and quasi-compact.

**Proof.**
Let $f : X \to S$ be affine. Quasi-compactness is immediate from Schemes, Lemma 26.19.2. We will show $f$ is separated using Schemes, Lemma 26.21.7. Let $x_1, x_2 \in X$ be points of $X$ which map to the same point $s \in S$. Choose any affine open $W \subset S$ containing $s$. By assumption $f^{-1}(W)$ is affine. Apply the lemma cited with $U = V = f^{-1}(W)$.
$\square$

Lemma 29.11.3. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

The morphism $f$ is affine.

There exists an affine open covering $S = \bigcup W_ j$ such that each $f^{-1}(W_ j)$ is affine.

There exists a quasi-coherent sheaf of $\mathcal{O}_ S$-algebras $\mathcal{A}$ and an isomorphism $X \cong \underline{\mathop{\mathrm{Spec}}}_ S(\mathcal{A})$ of schemes over $S$. See Constructions, Section 27.4 for notation.

Moreover, in this case $X = \underline{\mathop{\mathrm{Spec}}}_ S(f_*\mathcal{O}_ X)$.

**Proof.**
It is obvious that (1) implies (2).

Assume $S = \bigcup _{j \in J} W_ j$ is an affine open covering such that each $f^{-1}(W_ j)$ is affine. By Schemes, Lemma 26.19.2 we see that $f$ is quasi-compact. By Schemes, Lemma 26.21.6 we see the morphism $f$ is quasi-separated. Hence by Schemes, Lemma 26.24.1 the sheaf $\mathcal{A} = f_*\mathcal{O}_ X$ is a quasi-coherent sheaf of $\mathcal{O}_ S$-algebras. Thus we have the scheme $g : Y = \underline{\mathop{\mathrm{Spec}}}_ S(\mathcal{A}) \to S$ over $S$. The identity map $\text{id} : \mathcal{A} = f_*\mathcal{O}_ X \to f_*\mathcal{O}_ X$ provides, via the definition of the relative spectrum, a morphism $can : X \to Y$ over $S$, see Constructions, Lemma 27.4.7. By assumption and the lemma just cited the restriction $can|_{f^{-1}(W_ j)} : f^{-1}(W_ j) \to g^{-1}(W_ j)$ is an isomorphism. Thus $can$ is an isomorphism. We have shown that (2) implies (3).

Assume (3). By Constructions, Lemma 27.4.6 we see that the inverse image of every affine open is affine, and hence the morphism is affine by definition. $\square$

Remark 29.11.4. We can also argue directly that (2) implies (1) in Lemma 29.11.3 above as follows. Assume $S = \bigcup W_ j$ is an affine open covering such that each $f^{-1}(W_ j)$ is affine. First argue that $\mathcal{A} = f_*\mathcal{O}_ X$ is quasi-coherent as in the proof above. Let $\mathop{\mathrm{Spec}}(R) = V \subset S$ be affine open. We have to show that $f^{-1}(V)$ is affine. Set $A = \mathcal{A}(V) = f_*\mathcal{O}_ X(V) = \mathcal{O}_ X(f^{-1}(V))$. By Schemes, Lemma 26.6.4 there is a canonical morphism $\psi : f^{-1}(V) \to \mathop{\mathrm{Spec}}(A)$ over $\mathop{\mathrm{Spec}}(R) = V$. By Schemes, Lemma 26.11.6 there exists an integer $n \geq 0$, a standard open covering $V = \bigcup _{i = 1, \ldots , n} D(h_ i)$, $h_ i \in R$, and a map $a : \{ 1, \ldots , n\} \to J$ such that each $D(h_ i)$ is also a standard open of the affine scheme $W_{a(i)}$. The inverse image of a standard open under a morphism of affine schemes is standard open, see Algebra, Lemma 10.17.4. Hence we see that $f^{-1}(D(h_ i))$ is a standard open of $f^{-1}(W_{a(i)})$, in particular that $f^{-1}(D(h_ i))$ is affine. Because $\mathcal{A}$ is quasi-coherent we have $A_{h_ i} = \mathcal{A}(D(h_ i)) = \mathcal{O}_ X(f^{-1}(D(h_ i)))$, so $f^{-1}(D(h_ i))$ is the spectrum of $A_{h_ i}$. It follows that the morphism $\psi $ induces an isomorphism of the open $f^{-1}(D(h_ i))$ with the open $\mathop{\mathrm{Spec}}(A_{h_ i})$ of $\mathop{\mathrm{Spec}}(A)$. Since $f^{-1}(V) = \bigcup f^{-1}(D(h_ i))$ and $\mathop{\mathrm{Spec}}(A) = \bigcup \mathop{\mathrm{Spec}}(A_{h_ i})$ we win.

Lemma 29.11.5. Let $S$ be a scheme. There is an anti-equivalence of categories

which associates to $f : X \to S$ the sheaf $f_*\mathcal{O}_ X$. Moreover, this equivalence is compatible with arbitrary base change.

**Proof.**
The functor from right to left is given by $\underline{\mathop{\mathrm{Spec}}}_ S$. The two functors are mutually inverse by Lemma 29.11.3 and Constructions, Lemma 27.4.6 part (3). The final statement is Constructions, Lemma 27.4.6 part (2).
$\square$

Lemma 29.11.6. Let $f : X \to S$ be an affine morphism of schemes. Let $\mathcal{A} = f_*\mathcal{O}_ X$. The functor $\mathcal{F} \mapsto f_*\mathcal{F}$ induces an equivalence of categories

Moreover, an $\mathcal{A}$-module is quasi-coherent as an $\mathcal{O}_ S$-module if and only if it is quasi-coherent as an $\mathcal{A}$-module.

**Proof.**
Omitted.
$\square$

Lemma 29.11.7. The composition of affine morphisms is affine.

**Proof.**
Let $f : X \to Y$ and $g : Y \to Z$ be affine morphisms. Let $U \subset Z$ be affine open. Then $g^{-1}(U)$ is affine by assumption on $g$. Whereupon $f^{-1}(g^{-1}(U))$ is affine by assumption on $f$. Hence $(g \circ f)^{-1}(U)$ is affine.
$\square$

Lemma 29.11.8. The base change of an affine morphism is affine.

**Proof.**
Let $f : X \to S$ be an affine morphism. Let $S' \to S$ be any morphism. Denote $f' : X_{S'} = S' \times _ S X \to S'$ the base change of $f$. For every $s' \in S'$ there exists an open affine neighbourhood $s' \in V \subset S'$ which maps into some open affine $U \subset S$. By assumption $f^{-1}(U)$ is affine. By the material in Schemes, Section 26.17 we see that $f^{-1}(U)_ V = V \times _ U f^{-1}(U)$ is affine and equal to $(f')^{-1}(V)$. This proves that $S'$ has an open covering by affines whose inverse image under $f'$ is affine. We conclude by Lemma 29.11.3 above.
$\square$

Lemma 29.11.9. A closed immersion is affine.

**Proof.**
The first indication of this is Schemes, Lemma 26.8.2. See Schemes, Lemma 26.10.1 for a complete statement.
$\square$

Lemma 29.11.10. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Let $s \in \Gamma (X, \mathcal{L})$. The inclusion morphism $j : X_ s \to X$ is affine.

**Proof.**
This follows from Properties, Lemma 28.26.4 and the definition.
$\square$

Lemma 29.11.11. Suppose $g : X \to Y$ is a morphism of schemes over $S$.

If $X$ is affine over $S$ and $\Delta : Y \to Y \times _ S Y$ is affine, then $g$ is affine.

If $X$ is affine over $S$ and $Y$ is separated over $S$, then $g$ is affine.

A morphism from an affine scheme to a scheme with affine diagonal is affine.

A morphism from an affine scheme to a separated scheme is affine.

**Proof.**
Proof of (1). The base change $X \times _ S Y \to Y$ is affine by Lemma 29.11.8. The morphism $(1, g) : X \to X \times _ S Y$ is the base change of $Y \to Y \times _ S Y$ by the morphism $X \times _ S Y \to Y \times _ S Y$. Hence it is affine by Lemma 29.11.8. The composition of affine morphisms is affine (see Lemma 29.11.7) and (1) follows. Part (2) follows from (1) as a closed immersion is affine (see Lemma 29.11.9) and $Y/S$ separated means $\Delta $ is a closed immersion. Parts (3) and (4) are special cases of (1) and (2).
$\square$

Lemma 29.11.12. A morphism between affine schemes is affine.

**Proof.**
Immediate from Lemma 29.11.11 with $S = \mathop{\mathrm{Spec}}(\mathbf{Z})$. It also follows directly from the equivalence of (1) and (2) in Lemma 29.11.3.
$\square$

Lemma 29.11.13. Let $S$ be a scheme. Let $A$ be an Artinian ring. Any morphism $\mathop{\mathrm{Spec}}(A) \to S$ is affine.

**Proof.**
Omitted.
$\square$

Lemma 29.11.14. Let $j : Y \to X$ be an immersion of schemes. Assume there exists an open $U \subset X$ with complement $Z = X \setminus U$ such that

$U \to X$ is affine,

$j^{-1}(U) \to U$ is affine, and

$j(Y) \cap Z$ is closed.

Then $j$ is affine. In particular, if $X$ is affine, so is $Y$.

**Proof.**
By Schemes, Definition 26.10.2 there exists an open subscheme $W \subset X$ such that $j$ factors as a closed immersion $i : Y \to W$ followed by the inclusion morphism $W \to X$. Since a closed immersion is affine (Lemma 29.11.9), we see that for every $x \in W$ there is an affine open neighbourhood of $x$ in $X$ whose inverse image under $j$ is affine. If $x \in U$, then the same thing is true by assumption (2). Finally, assume $x \in Z$ and $x \not\in W$. Then $x \not\in j(Y) \cap Z$. By assumption (3) we can find an affine open neighbourhood $V \subset X$ of $x$ which does not meet $j(Y) \cap Z$. Then $j^{-1}(V) = j^{-1}(V \cap U)$ which is affine by assumptions (1) and (2). It follows that $j$ is affine by Lemma 29.11.3.
$\square$

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