The Stacks project

Lemma 15.11.5. Let $(A, I)$ be a pair. Let $A \to B$ be a finite type ring map such that $B/IB = C_1 \times C_2$ with $A/I \to C_1$ finite. Let $B'$ be the integral closure of $A$ in $B$. Then we can write $B'/IB' = C_1 \times C'_2$ such that the map $B'/IB' \to B/IB$ preserves product decompositions and there exists a $g \in B'$ mapping to $(1, 0)$ in $C_1 \times C'_2$ with $B'_ g \to B_ g$ an isomorphism.

Proof. Observe that $A \to B$ is quasi-finite at every prime of the closed subset $T = \mathop{\mathrm{Spec}}(C_1) \subset \mathop{\mathrm{Spec}}(B)$ (this follows by looking at fibre rings, see Algebra, Definition 10.122.3). Consider the diagram of topological spaces

\[ \xymatrix{ \mathop{\mathrm{Spec}}(B) \ar[rr]_\phi \ar[rd]_\psi & & \mathop{\mathrm{Spec}}(B') \ar[ld]^{\psi '} \\ & \mathop{\mathrm{Spec}}(A) } \]

By Algebra, Theorem 10.123.12 for every $\mathfrak p \in T$ there is a $h_\mathfrak p \in B'$, $h_\mathfrak p \not\in \mathfrak p$ such that $B'_ h \to B_ h$ is an isomorphism. The union $U = \bigcup D(h_\mathfrak p)$ gives an open $U \subset \mathop{\mathrm{Spec}}(B')$ such that $\phi ^{-1}(U) \to U$ is a homeomorphism and $T \subset \phi ^{-1}(U)$. Since $T$ is open in $\psi ^{-1}(V(I))$ we conclude that $\phi (T)$ is open in $U \cap (\psi ')^{-1}(V(I))$. Thus $\phi (T)$ is open in $(\psi ')^{-1}(V(I))$. On the other hand, since $C_1$ is finite over $A/I$ it is finite over $B'$. Hence $\phi (T)$ is a closed subset of $\mathop{\mathrm{Spec}}(B')$ by Algebra, Lemmas 10.41.6 and 10.36.22. We conclude that $\mathop{\mathrm{Spec}}(B'/IB') \supset \phi (T)$ is open and closed. By Algebra, Lemma 10.24.3 we get a corresponding product decomposition $B'/IB' = C'_1 \times C'_2$. The map $B'/IB' \to B/IB$ maps $C'_1$ into $C_1$ and $C'_2$ into $C_2$ as one sees by looking at what happens on spectra (hint: the inverse image of $\phi (T)$ is exactly $T$; some details omitted). Pick a $g \in B'$ mapping to $(1, 0)$ in $C'_1 \times C'_2$ such that $D(g) \subset U$; this is possible because $\mathop{\mathrm{Spec}}(C'_1)$ and $\mathop{\mathrm{Spec}}(C'_2)$ are disjoint and closed in $\mathop{\mathrm{Spec}}(B')$ and $\mathop{\mathrm{Spec}}(C'_1)$ is contained in $U$. Then $B'_ g \to B_ g$ defines a homeomorphism on spectra and an isomorphism on local rings (by our choice of $U$ above). Hence it is an isomorphism, as follows for example from Algebra, Lemma 10.23.1. Finally, it follows that $C'_1 = C_1$ and the proof is complete. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 15.11: Henselian pairs

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 09XH. Beware of the difference between the letter 'O' and the digit '0'.