What is an explicit example of a lifting in the valuative criterion for properness?

Question

The valuative criterion states that given a diagram $$ \begin{matrix} \text{Spec}(K) \to & X \\ \downarrow & \downarrow f \\ \text{Spec}(D) \to & Y \end{matrix} $$ where $K$ is a field and $D$ a valuation ring, there should be a lifting of $\text{Spec}(D) \to X$. What is an explicit example of this for $f:\mathbb{P}^n_\mathbb{Z} \to \mathbb{Z}$?

Answer 1

Here is a concrete example, inspired by Daniel's comment. We'll consider the projective scheme $\mathbb P_{\mathbb Z}^1$ (i.e. $n = 1$), and we'll take $K = \mathbb C(t)$ and $D = \mathbb C[t]_{(t)}$, with the morphism from ${\rm Spec \ } \mathbb C(t)$ to ${\rm Spec \ } \mathbb C[t]_{(t)}$ given by the natural ring morphism $\mathbb C[t]_{(t)} \hookrightarrow \mathbb C(t)$. We're given a morphism $\phi : {\rm Spec \ } \mathbb C(t) \to \mathbb P_{\mathbb Z}^1$ and we need to construct a morphism $\phi' : {\rm Spec \ } \mathbb C[t]_{(t)} \to \mathbb P_{\mathbb Z}^1$ such that your diagram commutes.

We can cover $\mathbb P_{\mathbb Z}^1 $ with two open affines, $\mathbb A_{\mathbb Z}^1 = {\rm Spec \ } \mathbb Z[u]$ and $ \mathbb A_{\mathbb Z}^1 = {\rm Spec \ } \mathbb Z [v]$, with $u = v^{-1}$ on their overlap. Since ${\rm Spec \ } \mathbb C (t)$ consists of a single point (namely, the generic point), the preimage of each of the two $\mathbb A_{\mathbb Z}^1$'s under $\phi : {\rm Spec \ } \mathbb C(t) \to \mathbb P_\mathbb Z^1$ is the whole of ${\rm Spec \ } \mathbb C(t)$. Let's assume that $\phi({\rm Spec \ }\mathbb C(t))$ intersects both copies of $\mathbb A_{\mathbb Z}^1$ (otherwise $\phi$ will map ${\rm Spec \ } \mathbb C(t)$ into a $\mathbb P_{\mathbb Z}^0 \subset \mathbb P_{\mathbb Z}^1$ and the story will be boring). Then the morphism $\phi$ is specified by a pair of $\mathbb Z$-algebra morphisms $f_1 : \mathbb Z[u] \to \mathbb C(t)$ and $f_2 : \mathbb Z[v] \to \mathbb C(t)$, obeying the relation $f_1(u) = (f_2(v))^{-1}$ for compatibility on the overlap.

Now let's write $f_1(u) = t^n h$, where $h$ is a unit in $\mathbb C[t]_{(t)}$ and $n$ is an integer. So $n$ is the valuation of $f_1(u)$. Then $f_2(v) = t^{-n}h^{-1}$. Clearly, at least one of $n$ or $-n$ is non-negative. So at least one of $f_1(u)$ and $f_2(v)$ is contained in $\mathbb C[t]_{(t)} \subset \mathbb C (t)$.

Without loss of generality, suppose that $f_1(u) \in \mathbb C[t]_{(t)} \subset \mathbb C(t) $. Since $f_1$ is a $\mathbb Z$-algebra morphism, it maps the whole of $\mathbb Z[u]$ into $\mathbb C[t]_{(t)}$. Using this morphism from $\mathbb Z[u]$ to $\mathbb C[t]_{(t)}$, we get a corresponding morphism of schemes (over ${\rm Spec \ } \mathbb Z$) from ${\rm Spec \ }\mathbb C[t]_{(t)}$ to the first copy of $\mathbb A_{\mathbb Z}^1$. Composing this with the natural open immersion $\mathbb A_{\mathbb Z}^1 \hookrightarrow \mathbb P_{\mathbb Z}^1$, we obtain a morphism $\phi ' $ from ${\rm Spec \ }\mathbb C[t]_{(t)}$ to $\mathbb P_{\mathbb Z}^1$.

It remains to check that the composition of the morphism ${\rm Spec \ }\mathbb C(t) \to {\rm Spec \ }\mathbb C[t]_{(t)}$ with our newly-constructed morphism $\phi ' : {\rm Spec \ }\mathbb C[t]_{(t)} \to \mathbb P_{\mathbb Z}^1$ is equal to the original morphism $\phi : {\rm Spec \ }\mathbb C(t) \to \mathbb P_{\mathbb Z}^1$. To check this, note that the images of both morphisms are contained inside the first $\mathbb A_{\mathbb Z}^1$, so we just need to check that the composition of the corresponding ring morphism $\mathbb Z[u] \to \mathbb C[t]_{(t)}$ with the inclusion morphism $\mathbb C[t]_{(t)} \hookrightarrow \mathbb C(t)$ is equal to the morphism $f_1 : \mathbb Z[u] \to \mathbb C(t)$. This is true by construction.

The nice thing about this example is that you can generalise it to any valuation domain $D$, and to $\mathbb P_{\mathbb Z}^n$ with $n$ greater than one. The argument will remain the same. Also notice how similar this is to the classical proof that any rational map from a smooth curve to a projective variety can be extended to a morphism: indeed, if you can think of $K$ as the function field of the smooth curve and $D$ as the local ring at a chosen point on the curve, then what we've done is essentially reproduce this classical argument.