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:

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

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

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 $g'(y') = y$,

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 $g'(x') = x$.
$\square$

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