## Tag `02JQ`

Chapter 28: Morphisms of Schemes > Section 28.6: Scheme theoretic image

Lemma 28.6.5. Let $f : X \to Y$ be a quasi-compact morphism. Let $Z$ be the scheme theoretic image of $f$. Let $z \in Z$. There exists a valuation ring $A$ with fraction field $K$ and a commutative diagram $$ \xymatrix{ \mathop{\rm Spec}(K) \ar[rr] \ar[d] & & X \ar[d] \ar[ld] \\ \mathop{\rm Spec}(A) \ar[r] & Z \ar[r] & Y } $$ such that the closed point of $\mathop{\rm Spec}(A)$ maps to $z$. In particular any point of $Z$ is the specialization of a point of $f(X)$.

Proof.Let $z \in \mathop{\rm Spec}(R) = V \subset Y$ be an affine open neighbourhood of $z$. By Lemma 28.6.3 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 $Y = \mathop{\rm Spec}(R)$ 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 $U_i = \mathop{\rm Spec}(A_i)$. Let $I = \mathop{\rm Ker}(R \to A_1 \times \ldots \times A_n)$. By Lemma 28.6.3 again we see that $Z$ corresponds to the closed subscheme $\mathop{\rm Spec}(R/I)$ 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 $R_{\mathfrak p} \to (A_1)_{\mathfrak p} \times \ldots \times (A_1)_{\mathfrak p}$ 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 10.49.2 we can choose a valuation ring $A \subset K = f.f.(A_i/\mathfrak q_i)$ dominating the local ring $R_{\mathfrak p}/\mathfrak p_iR_{\mathfrak p} \subset f.f.(A_i/\mathfrak q_i)$. This gives the desired diagram. Some details omitted. $\square$

The code snippet corresponding to this tag is a part of the file `morphisms.tex` and is located in lines 867–881 (see updates for more information).

```
\begin{lemma}
\label{lemma-reach-points-scheme-theoretic-image}
Let $f : X \to Y$ be a quasi-compact morphism.
Let $Z$ be the scheme theoretic image of $f$.
Let $z \in Z$. There exists a valuation ring $A$ with
fraction field $K$ and a commutative diagram
$$
\xymatrix{
\Spec(K) \ar[rr] \ar[d] & & X \ar[d] \ar[ld] \\
\Spec(A) \ar[r] & Z \ar[r] & Y
}
$$
such that the closed point of $\Spec(A)$ maps to $z$. In particular
any point of $Z$ is the specialization of a point of $f(X)$.
\end{lemma}
\begin{proof}
Let $z \in \Spec(R) = V \subset Y$ be an affine open
neighbourhood of $z$. By
Lemma \ref{lemma-quasi-compact-scheme-theoretic-image}
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 $Y = \Spec(R)$ 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 $U_i = \Spec(A_i)$.
Let $I = \Ker(R \to A_1 \times \ldots \times A_n)$.
By Lemma \ref{lemma-quasi-compact-scheme-theoretic-image}
again we see that $Z$ corresponds to the closed subscheme
$\Spec(R/I)$ 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 \ref{algebra-proposition-localization-exact})
we see that
$R_{\mathfrak p} \to
(A_1)_{\mathfrak p} \times \ldots \times (A_1)_{\mathfrak p}$
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 \ref{algebra-lemma-dominate} we can choose a valuation ring
$A \subset K = f.f.(A_i/\mathfrak q_i)$ dominating
the local ring
$R_{\mathfrak p}/\mathfrak p_iR_{\mathfrak p} \subset f.f.(A_i/\mathfrak q_i)$.
This gives the desired diagram. Some details omitted.
\end{proof}
```

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