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. $\square$
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