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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