The Stacks project

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

\[ T = \mathop{\mathrm{Spec}}(A[t, x_1, \ldots , x_ n]/(f_1 x_1 + \ldots + f_ n x_ n - 1)) \]

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

\[ \varphi : A'[y_1, \ldots , y_ n] \longrightarrow A'[t, x_1, \ldots , x_ n, x_{n + 1}]/(f'_1 x_1 + \ldots + f'_ n x_ n - 1) \]

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

\[ \psi : A'[t, x_1, \ldots , x_ n, x_{n + 1}]/(f'_1 x_1 + \ldots + f'_ n x_ n - 1) \longrightarrow A'[y_1, \ldots , y_ n] \]

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

\[ \begin{matrix} \sum f'_ i(y_ i + a'_ i - a'_ i(\sum f'_ j y_ j)) = (\sum f'_ iy_ i) + 1 - (\sum f'_ j y_ j) = 1 \end{matrix} \]

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

\[ \begin{matrix} \varphi (\psi (t) = \varphi (-1 + \sum f'_ i y_ i) = -1 + \sum f'_ i (a'_ i t + x_ i) = t \\ \varphi (\psi (x_ i)) = \varphi (y_ i + a'_ i - a'_ i(\sum f'_ j y_ j)) = a'_ i t + x_ i + a'_ i - a'_ i(\sum f'_ ja'_ jt + f'_ jx_ j) = x_ i \\ \psi (\varphi (y_ i)) = \psi (a'_ i t + x_ i) = a'_ i(-1 + \sum f'_ j y_ j) + y_ i + a'_ i - a'_ i(\sum f'_ j y_ j) = y_ i \end{matrix} \]

This finishes the proof. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 28.18: Quasi-affine schemes

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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0F82. Beware of the difference between the letter 'O' and the digit '0'.