Lemma 32.15.4. Let $f : X \to S$ be a finite type morphism of schemes. Assume $S$ is locally Noetherian. Then the following are equivalent

1. $f$ is universally closed,

2. for every $n$ the morphism $\mathbf{A}^ n \times X \to \mathbf{A}^ n \times S$ is closed,

3. for any diagram (32.15.1.1) there exists some dotted arrow,

4. for all diagrams (32.15.1.1) with $A$ a discrete valuation ring there exists some dotted arrow.

Proof. The equivalence of (1) and (2) is a special case of Lemma 32.14.2. The equivalence of (1) and (3) is a special case of Schemes, Proposition 26.20.6. Trivially (3) implies (4). Thus all we have to do is prove that (4) implies (2). We will prove that $\mathbf{A}^ n \times X \to \mathbf{A}^ n \times S$ is closed by the criterion of Schemes, Lemma 26.19.8. Pick $n$ and a specialization $z \leadsto z'$ of points in $\mathbf{A}^ n \times S$ and a point $y \in \mathbf{A}^ n \times X$ lying over $z$. Note that $\kappa (y)$ is a finitely generated field extension of $\kappa (z)$ as $\mathbf{A}^ n \times X \to \mathbf{A}^ n \times S$ is of finite type. Hence by Properties, Lemma 28.5.10 or Algebra, Lemma 10.119.13 implies that there exists a discrete valuation ring $A \subset \kappa (y)$ with fraction field $\kappa (z)$ dominating the image of $\mathcal{O}_{\mathbf{A}^ n \times S, z'}$ in $\kappa (z)$. This gives a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(\kappa (y)) \ar[r] \ar[d] & \mathbf{A}^ n \times X \ar[d] \ar[r] & X \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] & \mathbf{A}^ n \times S \ar[r] & S }$

Now property (4) implies that there exists a morphism $\mathop{\mathrm{Spec}}(A) \to X$ which fits into this diagram. Since we already have the morphism $\mathop{\mathrm{Spec}}(A) \to \mathbf{A}^ n$ from the left lower horizontal arrow we also get a morphism $\mathop{\mathrm{Spec}}(A) \to \mathbf{A}^ n \times X$ fitting into the left square. Thus the image $y' \in \mathbf{A}^ n \times X$ of the closed point is a specialization of $y$ lying over $z'$. This proves that specializations lift along $\mathbf{A}^ n \times X \to \mathbf{A}^ n \times S$ and we win. $\square$

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