Definition 28.18.1. A scheme $X$ is called *quasi-affine* if it is quasi-compact and isomorphic to an open subscheme of an affine scheme.

## 28.18 Quasi-affine schemes

Lemma 28.18.2. Let $A$ be a ring and let $U \subset \mathop{\mathrm{Spec}}(A)$ be a quasi-compact open subscheme. For $\mathcal{F}$ quasi-coherent on $U$ the canonical map

is an isomorphism.

**Proof.**
Denote $j : U \to \mathop{\mathrm{Spec}}(A)$ the inclusion morphism. Then $H^0(U, \mathcal{F}) = H^0(\mathop{\mathrm{Spec}}(A), j_*\mathcal{F})$ and $j_*\mathcal{F}$ is quasi-coherent by Schemes, Lemma 26.24.1. Hence $j_*\mathcal{F} = \widetilde{H^0(U, \mathcal{F})}$ by Schemes, Lemma 26.7.5. Restricting back to $U$ we get the lemma.
$\square$

Lemma 28.18.3. Let $X$ be a scheme. Let $f \in \Gamma (X, \mathcal{O}_ X)$. Assume $X$ is quasi-compact and quasi-separated and assume that $X_ f$ is affine. Then the canonical morphism

from Schemes, Lemma 26.6.4 induces an isomorphism of $X_ f = j^{-1}(D(f))$ onto the standard affine open $D(f) \subset \mathop{\mathrm{Spec}}(\Gamma (X, \mathcal{O}_ X))$.

**Proof.**
This is clear as $j$ induces an isomorphism of rings $\Gamma (X, \mathcal{O}_ X)_ f \to \mathcal{O}_ X(X_ f)$ by Lemma 28.17.1 above.
$\square$

Lemma 28.18.4. Let $X$ be a scheme. Then $X$ is quasi-affine if and only if the canonical morphism

from Schemes, Lemma 26.6.4 is a quasi-compact open immersion.

**Proof.**
If the displayed morphism is a quasi-compact open immersion then $X$ is isomorphic to a quasi-compact open subscheme of $\mathop{\mathrm{Spec}}(\Gamma (X, \mathcal{O}_ X))$ and clearly $X$ is quasi-affine.

Assume $X$ is quasi-affine, say $X \subset \mathop{\mathrm{Spec}}(R)$ is quasi-compact open. This in particular implies that $X$ is separated, see Schemes, Lemma 26.23.9. Let $A = \Gamma (X, \mathcal{O}_ X)$. Consider the ring map $R \to A$ coming from $R = \Gamma (\mathop{\mathrm{Spec}}(R), \mathcal{O}_{\mathop{\mathrm{Spec}}(R)})$ and the restriction mapping of the sheaf $\mathcal{O}_{\mathop{\mathrm{Spec}}(R)}$. By Schemes, Lemma 26.6.4 we obtain a factorization:

of the inclusion morphism. Let $x \in X$. Choose $r \in R$ such that $x \in D(r)$ and $D(r) \subset X$. Denote $f \in A$ the image of $r$ in $A$. The open $X_ f$ of Lemma 28.17.1 above is equal to $D(r) \subset X$ and hence $A_ f \cong R_ r$ by the conclusion of that lemma. Hence $D(r) \to \mathop{\mathrm{Spec}}(A)$ is an isomorphism onto the standard affine open $D(f)$ of $\mathop{\mathrm{Spec}}(A)$. Since $X$ can be covered by such affine opens $D(f)$ we win. $\square$

Lemma 28.18.5. Let $U \to V$ be an open immersion of quasi-affine schemes. Then

is cartesian.

**Proof.**
The diagram is commutative by Schemes, Lemma 26.6.4. Write $A = \Gamma (U, \mathcal{O}_ U)$ and $B = \Gamma (V, \mathcal{O}_ V)$. Let $g \in B$ be such that $V_ g$ is affine and contained in $U$. This means that if $f$ is the image of $g$ in $A$, then $U_ f = V_ g$. By Lemma 28.18.3 we see that $j'$ induces an isomorphism of $V_ g$ with the standard open $D(g)$ of $\mathop{\mathrm{Spec}}(B)$. Thus $V_ g \times _{\mathop{\mathrm{Spec}}(B)} \mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(A)$ is an isomorphism onto $D(f) \subset \mathop{\mathrm{Spec}}(A)$. By Lemma 28.18.3 again $j$ maps $U_ f$ isomorphically to $D(f)$. Thus we see that $U_ f = U_ f \times _{\mathop{\mathrm{Spec}}(B)} \mathop{\mathrm{Spec}}(A)$. Since by Lemma 28.18.4 we can cover $U$ by $V_ g = U_ f$ as above, we see that $U \to U \times _{\mathop{\mathrm{Spec}}(B)} \mathop{\mathrm{Spec}}(A)$ is an isomorphism.
$\square$

Lemma 28.18.6. Let $X$ be a quasi-affine scheme. There exists an integer $n \geq 0$, an affine scheme $T$, and a morphism $T \to X$ such that for every morphism $X' \to X$ with $X'$ affine the fibre product $X' \times _ X T$ is isomorphic to $\mathbf{A}^ n_{X'}$ over $X'$.

**Proof.**
By definition, there exists a ring $A$ such that $X$ is isomorphic to a quasi-compact open subscheme $U \subset \mathop{\mathrm{Spec}}(A)$. Recall that the standard opens $D(f) \subset \mathop{\mathrm{Spec}}(A)$ form a basis for the topology, see Algebra, Section 10.17. Since $U$ is quasi-compact we can choose $f_1, \ldots , f_ n \in A$ such that $U = D(f_1) \cup \ldots \cup D(f_ n)$. Thus we may assume $X = \mathop{\mathrm{Spec}}(A) \setminus V(I)$ where $I = (f_1, \ldots , f_ n)$. We set

The structure morphism $T \to \mathop{\mathrm{Spec}}(A)$ factors through the open $X$ to give the morphism $T \to X$. If $X' = \mathop{\mathrm{Spec}}(A')$ and the morphism $X' \to X$ corresponds to the ring map $A \to A'$, then the images $f'_1, \ldots , f'_ n \in A'$ of $f_1, \ldots , f_ n$ generate the unit ideal in $A'$. Say $1 = f'_1 a'_1 + \ldots + f'_ n a'_ n$. The base change $X' \times _ X T$ is the spectrum of $A'[t, x_1, \ldots , x_ n]/(f'_1 x_1 + \ldots + f'_ n x_ n - 1)$. We claim the $A'$-algebra homomorphism

sending $y_ i$ to $a'_ i t + x_ i$ is an isomorphism. The claim finishes the proof of the lemma. The inverse of $\varphi $ is given by the $A'$-algebra homomorphism

sending $t$ to $-1 + f'_1 y_1 + \ldots + f'_ n y_ n$ and $x_ i$ to $y_ i + a'_ i - a'_ i(f'_1 y_1 + \ldots + f'_ n y_ n)$ for $i = 1, \ldots , n$. This makes sense because $\sum f'_ ix_ i$ is mapped to

To see the maps are mutually inverse one computes as follows:

This finishes the proof. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (2)

Comment #2046 by alex on

Comment #2047 by alex on