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Tag 0ARY

Lemma 27.18.4. Let $U \to V$ be an open immersion of quasi-affine schemes. Then $$\xymatrix{ U \ar[d] \ar[rr]_-j & & \mathop{\rm Spec}(\Gamma(U, \mathcal{O}_U)) \ar[d] \\ U \ar[r] & V \ar[r]^-{j'} & \mathop{\rm Spec}(\Gamma(V, \mathcal{O}_V)) }$$ is cartesian.

Proof. The diagram is commutative by Schemes, Lemma 25.6.4. 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 27.18.2 we see that $j'$ induces an isomorphism of $V_g$ with the standard open $D(g)$ of $\mathop{\rm Spec}(B)$. Thus $V_g \times_{\mathop{\rm Spec}(B)} \mathop{\rm Spec}(A) \to \mathop{\rm Spec}(A)$ is an isomorphism onto $D(f) \subset \mathop{\rm Spec}(A)$. By Lemma 27.18.2 again $j$ maps $U_f$ isomorphically to $D(f)$. Thus we see that $U_f = U_f \times_{\mathop{\rm Spec}(B)} \mathop{\rm Spec}(A)$. Since by Lemma 27.18.3 we can cover $U$ by $V_g = U_f$ as above, we see that $U \to U \times_{\mathop{\rm Spec}(B)} \mathop{\rm Spec}(A)$ is an isomorphism. $\square$

The code snippet corresponding to this tag is a part of the file properties.tex and is located in lines 2244–2254 (see updates for more information).

\begin{lemma}
\label{lemma-cartesian-diagram-quasi-affine}
Let $U \to V$ be an open immersion of quasi-affine schemes. Then
$$\xymatrix{ U \ar[d] \ar[rr]_-j & & \Spec(\Gamma(U, \mathcal{O}_U)) \ar[d] \\ U \ar[r] & V \ar[r]^-{j'} & \Spec(\Gamma(V, \mathcal{O}_V)) }$$
is cartesian.
\end{lemma}

\begin{proof}
The diagram is commutative by Schemes, Lemma
\ref{schemes-lemma-morphism-into-affine}.
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
\ref{lemma-invert-f-affine} we see that $j'$ induces an isomorphism of
$V_g$ with the standard open $D(g)$ of $\Spec(B)$. Thus
$V_g \times_{\Spec(B)} \Spec(A) \to \Spec(A)$ is an
isomorphism onto $D(f) \subset \Spec(A)$. By Lemma \ref{lemma-invert-f-affine}
again $j$ maps $U_f$ isomorphically to $D(f)$. Thus we see that
$U_f = U_f \times_{\Spec(B)} \Spec(A)$. Since by
Lemma \ref{lemma-quasi-affine} we can cover $U$ by $V_g = U_f$ as above,
we see that $U \to U \times_{\Spec(B)} \Spec(A)$ is an isomorphism.
\end{proof}

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