The Stacks project

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 $u' \not\in V(I)$. $\square$

Comments (3)

Comment #5899 by Thibaud van den Hove on

In the second paragraph, I think you want to distinguish between whether or not instead of whether or not : the inverse image in of a prime ideal in will always contain for all . Similarly, you also want instead of .

I also found a few typo's:

In the second paragraph, you want to consider as an ideal of and instead of and . The following are all in the last paragraph: * you wrote , and forgot the (the should be , but I had some trouble formatting, sorry about that), * you want instead of , * you wrote instead of , * you wrote instead of , * you use both and , but they should mean the same thing.

Comment #5900 by Thibaud van den Hove on

I tried putting the typo's in a list, but it looks like I has some more formatting problems, sorry about this.

Comment #6101 by on

Thanks very much. This can still be cleaned up further but I am going to take your comments as essentially telling me the proof isn't wrong. Fixes are here.

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 0B7J. Beware of the difference between the letter 'O' and the digit '0'.