The Stacks project

This can be found in the proof of [IV Proposition 12.1.1.5, EGA]

Lemma 37.18.2. Let $f : X \to S$ be a morphism of schemes. Let $y \in X$ be a point with image $t \in S$. Denote $Y \subset X$ the closure of $\{ y\} $ viewed as an integral closed subscheme of $X$. Let $s \in S$ and let $x \in Y_ s$ be a generic point of an irreducible component of $Y_ s$. There exists a cartesian diagram

\[ \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ S' \ar[r]^ g & S } \]

with the following properties:

  1. $S'$ is the spectrum of a valuation ring with generic point $t'$ and closed point $s'$,

  2. $g(t') = t$ and $g(s') = s$,

  3. there exists a point $y' \in X'_{t'}$ which is a generic point of an irreducible component of $(S' \times _ S Y)_{t'} = Y_ t \times _ t t'$ and satisfies $g'(y') = y$,

  4. denoting $Y' \subset X'$ the closure of $\{ y'\} $ viewed as an integral closed subscheme of $X'$ there exists a point $x' \in Y'_{s'}$ which is a generic point of an irreducible component of $Y'_{s'}$ with $g'(x') = x$.

Proof. We choose a valuation ring $R$, we set $S' = \mathop{\mathrm{Spec}}(R)$ with generic point $t'$ and closed point $s'$, and we choose a morphism $h : S' \to X$ with $h(t') = y$ and $h(s') = x$. See Schemes, Lemma 26.20.4. Set $g = f \circ h$ so that $g(t') = t$ and $g(s') = s$. Consider the base change

\[ \xymatrix{ X' \ar[r]_{g'} \ar[d] & X \ar[d] \\ S' \ar@/^1em/[u]^\sigma \ar[r]^-g & S } \]

We obtain a section $\sigma $ of the base change such that $h = g' \circ \sigma $.

Of course $\sigma $ factors through the base change $S' \times _ S Y$ of $Y$ as $h$ factors through $Y$. Let $y' \in X'_{t'} \subset X'$ be the generic point of an irreducible component of the fibre

\[ (S' \times _ S Y)_{t'} = Y_ t \times _ t t' \]

containing the point $\sigma (t')$, i.e., such that $y' \leadsto \sigma (t')$. Since $g'(y') \in Y_ t$ and $g(y') \leadsto g(\sigma (t')) = y$ we find that $g'(y') = y$ because $y$ is the generic point of the fibre $Y_ t$. Denote $Y' \subset X'$ the closure of $\{ y'\} $ in $X'$ viewed as an integral closed subscheme. Then $\sigma $ factors through $Y'$ as $\sigma (t') \in Y'$. Choose a generic point $x' \in Y'_{s'}$ of an irreducible component of $Y'_{s'}$ which contains $\sigma (s')$, i.e., we get $x' \leadsto \sigma (s')$ and hence $g'(x') \leadsto g'(\sigma (s')) = x$. Again as $x$ is a generic point of an irreducible component of $Y_ s$ by assumption and as $g'(Y') \subset Y$ we conclude that $g'(x') = x$. $\square$


Comments (0)


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