Lemma 15.6.2. In Situation 15.6.1 we have

as topological spaces.

Lemma 15.6.2. In Situation 15.6.1 we have

\[ \mathop{\mathrm{Spec}}(B') = \mathop{\mathrm{Spec}}(B) \amalg _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A') \]

as topological spaces.

**Proof.**
Since $B' = B \times _ A A'$ we obtain a commutative square of spectra, which induces a continuous map

\[ can : \mathop{\mathrm{Spec}}(B) \amalg _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A') \longrightarrow \mathop{\mathrm{Spec}}(B') \]

as the source is a pushout in the category of topological spaces (which exists by Topology, Section 5.29).

To show the map $can$ is surjective, let $\mathfrak q' \subset B'$ be a prime ideal. If $I \subset \mathfrak q'$ (here and below we take the liberty of considering $I$ as an ideal of $B'$ as well as an ideal of $A'$), then $\mathfrak q'$ corresponds to a prime ideal of $B$ and is in the image. If not, then pick $h \in I$, $h \not\in \mathfrak q'$. In this case $B_ h = A_ h = 0$ and the ring map $B'_ h \to A'_ h$ is an isomorphism, see Lemma 15.5.3. Thus we see that $\mathfrak q'$ corresponds to a unique prime ideal $\mathfrak p' \subset A'$ which does not contain $I$.

Since $B' \to B$ is surjective, we see that $can$ is injective on the summand $\mathop{\mathrm{Spec}}(B)$. We have seen above that $\mathop{\mathrm{Spec}}(A') \to \mathop{\mathrm{Spec}}(B')$ is injective on the complement of $V(I) \subset \mathop{\mathrm{Spec}}(A')$. Since $V(I) \subset \mathop{\mathrm{Spec}}(A')$ is exactly the image of $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(A')$ a trivial set theoretic argument shows that $can$ is injective.

To finish the proof we have to show that $can$ is open. To do this, observe that an open of the pushout is of the form $V \amalg U'$ where $V \subset \mathop{\mathrm{Spec}}(B)$ and $U' \subset \mathop{\mathrm{Spec}}(A')$ are opens whose inverse images in $\mathop{\mathrm{Spec}}(A)$ agree. Let $v \in V$. We can find a $g \in B$ such that $v \in D(g) \subset V$. Let $f \in A$ be the image. Pick $f' \in A'$ mapping to $f$. Then $D(f') \cap U' \cap V(I) = D(f') \cap V(I)$. Hence $V(I) \cap D(f')$ and $D(f') \cap (U')^ c$ are disjoint closed subsets of $D(f') = \mathop{\mathrm{Spec}}(A'_{f'})$. Write $(U')^ c = V(J)$ for some ideal $J \subset A'$. Since $A'_{f'} \to A'_{f'}/IA'_{f'} \times A'_{f'}/JA'_{f'}$ is surjective by the disjointness just shown, we can find an $a'' \in A'_{f'}$ mapping to $1$ in $A'_{f'}/IA'_{f'}$ and mapping to zero in $A'_{f'}/JA'_{f'}$. Clearing denominators, we find an element $a' \in J$ mapping to $f^ n$ in $A$. Then $D(a'f') \subset U'$. Let $h' = (g^{n + 1}, a'f') \in B'$. Since $B'_{h'} = B_{g^{n + 1}} \times _{A_{f^{n + 1}}} A'_{a'f'}$ by a previously cited lemma, we see that $D(h')$ pulls back to an open neighbourhood of $v$ in the pushout, i.e., the image of $V \amalg U'$ contains an open neighbourhood of the image of $v$. We omit the (easier) proof that the same thing is true for $u' \in U'$ with $u' \not\in V(I)$. $\square$

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## Comments (3)

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