Lemma 29.6.5. Let $f : X \to Y$ be a quasi-compact morphism. Let $Z$ be the scheme theoretic image of $f$. Let $z \in Z$^{1}. There exists a valuation ring $A$ with fraction field $K$ and a commutative diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[rr] \ar[d] & & X \ar[d] \ar[ld] \\ \mathop{\mathrm{Spec}}(A) \ar[r] & Z \ar[r] & Y } \]

such that the closed point of $\mathop{\mathrm{Spec}}(A)$ maps to $z$. In particular any point of $Z$ is the specialization of a point of $f(X)$.

**Proof.**
Let $z \in \mathop{\mathrm{Spec}}(R) = V \subset Y$ be an affine open neighbourhood of $z$. By Lemma 29.6.3 the intersection $Z \cap V$ is the scheme theoretic image of $f^{-1}(V) \to V$. Hence we may replace $Y$ by $V$ and assume $Y = \mathop{\mathrm{Spec}}(R)$ is affine. In this case $X$ is quasi-compact as $f$ is quasi-compact. Say $X = U_1 \cup \ldots \cup U_ n$ is a finite affine open covering. Write $U_ i = \mathop{\mathrm{Spec}}(A_ i)$. Let $I = \mathop{\mathrm{Ker}}(R \to A_1 \times \ldots \times A_ n)$. By Lemma 29.6.3 again we see that $Z$ corresponds to the closed subscheme $\mathop{\mathrm{Spec}}(R/I)$ of $Y$. If $\mathfrak p \subset R$ is the prime corresponding to $z$, then we see that $I_{\mathfrak p} \subset R_{\mathfrak p}$ is not an equality. Hence (as localization is exact, see Algebra, Proposition 10.9.12) we see that $R_{\mathfrak p} \to (A_1)_{\mathfrak p} \times \ldots \times (A_ n)_{\mathfrak p}$ is not zero. Hence one of the rings $(A_ i)_{\mathfrak p}$ is not zero. Hence there exists an $i$ and a prime $\mathfrak q_ i \subset A_ i$ lying over a prime $\mathfrak p_ i \subset \mathfrak p$. By Algebra, Lemma 10.50.2 we can choose a valuation ring $A \subset K = \kappa (\mathfrak q_ i)$ dominating the local ring $R_{\mathfrak p}/\mathfrak p_ iR_{\mathfrak p} \subset \kappa (\mathfrak q_ i)$. This gives the desired diagram. Some details omitted.
$\square$

## Comments (5)

Comment #2762 by BCnrd on

Comment #2765 by Dario Weißmann on

Comment #2766 by BCnrd on

Comment #2767 by sdf on

Comment #2875 by Johan on

There are also: