1. Localization at a prime in a Dedekind domain

Let $A$ be a Dedekind domain (e.g., $\mathbb{Z}$ or $k[t]$) and $\mathfrak{p}$ a nonzero prime ideal. Then $A_{\mathfrak{p}}$ is a DVR.

Example: $R = \mathbb{Z}_{(p)}$ is the localization of $\mathbb{Z}$ at the prime $(p)$: $\mathbb{Z}_{(p)} = \left\{ \frac{a}{b} : a, b \in \mathbb{Z}, \gcd(b,p) = 1 \right\}$

The maximal ideal is $\mathfrak{m} = p\mathbb{Z}_{(p)}$, the field of fractions is $K = \mathbb{Q}$, and the residue field is $\kappa = \mathbb{F}_p$.

2. Power series ring

$R = k[[t]]$, the ring of formal power series over a field $k$, is a DVR with:

• Maximal ideal $\mathfrak{m} = (t)$
• Field of fractions $K = k((t))$ (Laurent series)
• Residue field $\kappa = k$

Every element can be written uniquely as $f = u t^n$ where $u \in k[[t]]^*$ (units) and $n \geq 0$.

3. P-adic integers

$R = \mathbb{Z}_p$, the ring of $p$-adic integers, is the completion of $\mathbb{Z}_{(p)}$ with respect to the $(p)$-adic topology: $\mathbb{Z}_p = \varprojlim \mathbb{Z}/p^n\mathbb{Z}$

It has:

• Maximal ideal $\mathfrak{m} = p\mathbb{Z}_p$
• Field of fractions $K = \mathbb{Q}_p$ (p-adic numbers)
• Residue field $\kappa = \mathbb{F}_p$

4. Localization of a polynomial ring

Let $R = k[x,y]_{(x,y)}/(xy - t)$ where $t \in k$. If $t \neq 0$, this is not a DVR, but if we take the completion at the origin, we get a DVR in many cases.

Actually, a cleaner example: $R = k[x]_{(x-a)}$ for any $a \in k$ is a DVR with uniformizer $\pi = x - a$.

5. Valuation ring of a discrete valuation

Given any discrete valuation $v: K^* \to \mathbb{Z}$ on a field $K$, the valuation ring $R = \{ x \in K : v(x) \geq 0 \} \cup \{0\}$ is a DVR with maximal ideal $\mathfrak{m} = \{ x \in K : v(x) > 0 \} \cup \{0\}$.

From a geometric perspective, $\text{Spec } R$ for a DVR $R$ represents a "germ of a curve with a marked point":

• $\text{Spec } K$ (generic point) represents the "curve minus a point"
• $\text{Spec } \kappa$ (closed point) represents the "missing point"

The inclusion $K \hookrightarrow R$ corresponds to the morphism $\text{Spec } R \to \text{Spec } K$, which is the inclusion of the generic point.

Concrete example: Consider $R = k[[t]]$.

• $\text{Spec } k[[t]]$ has two points: generic point $\eta$ (corresponding to $(0)$) and closed point $s$ (corresponding to $(t)$)
• $\text{Spec } k((t))$ has one point (the generic point)
• The morphism $\text{Spec } k[[t]] \to \text{Spec } k((t))$ includes the closed point into the generic point

Think of $\text{Spec } k[[t]]$ as a formal neighborhood of a point on a curve.

Consider $X = \mathbb{A}^1_k = \text{Spec } k[x]$ and $R = k[[t]]$.

Extending a generic point: A morphism $\text{Spec } k((t)) \to \mathbb{A}^1_k$ is given by specifying $x \in k((t))$.

Suppose $x = t^{-1} + t + t^2 + \cdots$ (a Laurent series with a pole at $t = 0$).

Can this extend to $\text{Spec } k[[t]] \to \mathbb{A}^1_k$?

An extension would require $x \in k[[t]]$ (a power series), but $x$ has a pole, so no extension exists.

This shows that $\mathbb{A}^1$ is not proper: there are curves (like $\text{Spec } k[[t]]$) and rational points on them that don't extend to the whole curve.

Another example: If $x = t + t^2 + \cdots \in k[[t]]$, then the morphism $\text{Spec } k((t)) \to \mathbb{A}^1$ extends uniquely to $\text{Spec } k[[t]] \to \mathbb{A}^1$ by the same formula.

This shows that $\mathbb{A}^1$ is separated: when an extension exists, it is unique.

Consider $X = \mathbb{P}^1_k$ with standard open cover $U_0 = \text{Spec } k[x]$ and $U_1 = \text{Spec } k[y]$ where $xy = 1$ on the overlap.

Extending a generic point with a pole: Let $R = k[[t]]$ and consider the morphism $\text{Spec } k((t)) \to \mathbb{P}^1$ given by $x = t^{-1}$.

In the chart $U_0$, we have $x = t^{-1} \notin k[[t]]$, so the generic point doesn't map into $U_0 \cap \text{Spec } k[[t]]$.

In the chart $U_1$, we have $y = 1/x = t \in k[[t]]$, so the map does extend to $\text{Spec } k[[t]] \to U_1 \subset \mathbb{P}^1$.

The extension is given by $[x:y] = [1:t]$ on $\text{Spec } k[[t]]$, which evaluates to:

• At the generic point: $[1:t] = [t^{-1}:1]$ (since we're in $k((t))$)
• At the closed point: $[1:0]$ (the "point at infinity")

This shows that $\mathbb{P}^1$ is proper: every rational point extends uniquely to the whole curve. The "missing point" at infinity provides a place for poles to land.

Let $X = \mathbb{A}^1_k = \text{Spec } k[x]$ over $S = \text{Spec } k$.

Claim: $\mathbb{A}^1_k$ is separated over $k$.

Proof via valuative criterion: Let $R$ be a DVR with field of fractions $K$, and suppose we have:

• A morphism $\text{Spec } K \to \mathbb{A}^1_k$ given by $x \mapsto \alpha \in K$
• Two extensions $\varphi_1, \varphi_2: \text{Spec } R \to \mathbb{A}^1_k$

Each extension $\varphi_i$ is determined by where $x$ maps, say $x \mapsto a_i \in R$.

For the extensions to agree on $\text{Spec } K$, we need: $a_1|_K = a_2|_K = \alpha$

But $R \hookrightarrow K$ is an inclusion, so $a_1 = a_2 = \alpha$. Since $\alpha \in K$ must actually be in $R$ for the extension to exist, we have $\alpha \in R$.

Therefore, if two extensions exist, they must be the same: $\varphi_1 = \varphi_2$.

This proves $\mathbb{A}^1_k$ is separated.

Consider the affine line with a doubled origin: let $X$ be obtained by gluing two copies of $\mathbb{A}^1_k$ along $\mathbb{A}^1 \setminus \{0\}$.

More precisely:

• Let $U_1 = \text{Spec } k[x]$ and $U_2 = \text{Spec } k[y]$
• Glue them along $D(x) \cong D(y) = \mathbb{A}^1 \setminus \{0\}$ via the identity $x = y$

The resulting scheme $X$ has two origins: $0_1 \in U_1$ and $0_2 \in U_2$.

Why it fails separatedness: Let $R = k[[t]]$ with field of fractions $K = k((t))$.

Consider the morphism $\text{Spec } K \to X$ given by $x = y = t \in k((t))$. This maps into the complement of both origins (since $t \neq 0$ in $K$).

This morphism has two distinct extensions to $\text{Spec } R \to X$:

1. $\varphi_1: \text{Spec } k[[t]] \to U_1$ given by $x = t$
2. $\varphi_2: \text{Spec } k[[t]] \to U_2$ given by $y = t$

Both extend the same generic point, but:

• $\varphi_1$ maps the closed point to $0_1$
• $\varphi_2$ maps the closed point to $0_2$

Since $0_1 \neq 0_2$, we have $\varphi_1 \neq \varphi_2$.

This violates the valuative criterion, so $X$ is not separated.

Separated gluing: Gluing two copies of $\mathbb{A}^1$ along a principal open to get $\mathbb{A}^1$ is separated.

Let $U_1 = \text{Spec } k[x]$ and $U_2 = \text{Spec } k[y]$, glued along $D(x) = D(y)$ via $y = x^{-1}$.

This gives $\mathbb{P}^1$, which is separated (in fact, proper).

Non-separated gluing: Gluing two copies of $\mathbb{A}^1$ by identifying the complement of the origin gives the line with doubled origin (previous example), which is not separated.

The key difference: in the separated case, the overlap is a principal open subset (corresponding to a localization), while in the non-separated case, the overlap is the complement of a closed point (which is not affine).

Let $X = \mathbb{P}^1_k$ over $S = \text{Spec } k$.

Claim: $\mathbb{P}^1_k$ is proper over $k$.

Proof via valuative criterion: Let $R$ be a DVR with field of fractions $K$, and let $\text{Spec } K \to \mathbb{P}^1_k$ be a morphism.

The morphism is given by homogeneous coordinates $[a_0 : a_1]$ with $a_0, a_1 \in K$, not both zero.

Case 1: $a_0 \neq 0$. Then we can write $[a_0 : a_1] = [1 : a_1/a_0]$. Let $v$ be the valuation on $K$ associated to $R$.

• If $v(a_1/a_0) \geq 0$, then $a_1/a_0 \in R$, so the morphism extends to $\text{Spec } R \to U_0$ (where $U_0 = \text{Spec } k[x_1/x_0]$) via $[1 : a_1/a_0]$.
• If $v(a_1/a_0) < 0$, then $v(a_0/a_1) > 0$, so $a_0/a_1 \in R$. The morphism extends to $\text{Spec } R \to U_1$ (where $U_1 = \text{Spec } k[x_0/x_1]$) via $[a_0/a_1 : 1]$.

Case 2: $a_0 = 0$, so $a_1 \neq 0$. Then $[0 : a_1] = [0 : 1]$ and the morphism extends to $\text{Spec } R \to U_1$ via $[0 : 1]$.

In all cases, an extension exists.

Uniqueness: Suppose we have two extensions $\varphi_1, \varphi_2: \text{Spec } R \to \mathbb{P}^1$.

Both are determined by homogeneous coordinates with coefficients in $R$. Since they agree on the generic point $\text{Spec } K$, and $R \to K$ is injective, they must agree on $\text{Spec } R$.

Therefore, $\mathbb{P}^1_k$ is proper.

Let $X = \mathbb{A}^1_k = \text{Spec } k[x]$ over $S = \text{Spec } k$.

Why it's not proper: Let $R = k[[t]]$ with $K = k((t))$.

Consider the morphism $\text{Spec } K \to \mathbb{A}^1_k$ given by $x = t^{-1}$.

To extend this to $\text{Spec } R \to \mathbb{A}^1_k$, we would need $x \in k[[t]]$. But $t^{-1} \notin k[[t]]$ (it has a pole at $t = 0$).

Therefore, no extension exists, violating the valuative criterion for properness.

The geometric intuition: $\mathbb{A}^1$ has "a point missing at infinity," and the curve $x = t^{-1}$ "escapes to infinity" as $t \to 0$.

Let $X = \mathbb{A}^1_k \setminus \{0\} = \text{Spec } k[x, x^{-1}]$ over $S = \text{Spec } k$.

Why it's not proper: Let $R = k[[t]]$ with $K = k((t))$.

Consider the morphism $\text{Spec } K \to X$ given by $x = t$.

To extend this to $\text{Spec } R \to X$, we would need $x \in k[[t]]$ and $x^{-1} \in k[[t]]$.

We have $x = t \in k[[t]]$, but $x^{-1} = t^{-1} \notin k[[t]]$.

Therefore, no extension exists.

The geometric intuition: the curve $x = t$ approaches the missing point $\{0\}$ as $t \to 0$, but there's no point there to extend to.

More generally, $\mathbb{P}^n_k$ is proper over $k$ for any $n \geq 0$.

Proof sketch: Let $R$ be a DVR with field of fractions $K$, and let $\text{Spec } K \to \mathbb{P}^n_k$ be a morphism given by $[a_0 : \cdots : a_n]$ with $a_i \in K$, not all zero.

Choose $i$ such that $v(a_i)$ is minimal (where $v$ is the valuation on $K$). Then $v(a_j/a_i) \geq 0$ for all $j$, so $a_j/a_i \in R$ for all $j$.

The morphism extends to $\text{Spec } R \to U_i \subset \mathbb{P}^n$ (where $U_i = \text{Spec } k[x_0/x_i, \ldots, \widehat{x_i/x_i}, \ldots, x_n/x_i]$) via $[a_0/a_i : \cdots : 1 : \cdots : a_n/a_i]$.

Uniqueness follows from the injectivity of $R \to K$ and the fact that homogeneous coordinates are unique up to scaling by units.

Let $X = \text{Bl}_0 \mathbb{A}^2_k$ be the blowup of $\mathbb{A}^2_k$ at the origin.

Recall that $X$ can be covered by two affine charts:

• $U_1 = \text{Spec } k[x, y_1]$ where $y_1 = y/x$
• $U_2 = \text{Spec } k[x_2, y]$ where $x_2 = x/y$

The exceptional divisor $E$ is given by $x = 0$ in $U_1$ and $y = 0$ in $U_2$, and is isomorphic to $\mathbb{P}^1$.

Claim: $X \to \text{Spec } k$ is proper.

Proof via valuative criterion: Let $R$ be a DVR with $K = \text{Frac}(R)$, and let $\text{Spec } K \to X$ be a morphism.

The morphism to $X$ corresponds to a morphism to $\mathbb{A}^2_k$ that is an isomorphism away from the origin, so it's given by $(x, y) \in K^2$, not both zero (after blowing up).

• If $v(x/y) \geq 0$, then $y_1 = y/x \in R$, and the morphism extends to $U_1$ via $(x, y_1)$.
• If $v(y/x) \geq 0$, then $x_2 = x/y \in R$, and the morphism extends to $U_2$ via $(x_2, y)$.

At least one of these conditions holds (since $v(x/y) + v(y/x) = 0$), so an extension exists.

Uniqueness follows from separatedness of the blowup (which can be checked directly or follows from the fact that the blowup is a projective morphism).

An important consequence of the valuative criterion:

Theorem: If $f: X \to S$ is proper with $S$ affine, then the fibers of $f$ are finite sets (as topological spaces). If additionally $X$ is affine, then $f$ is finite (i.e., $X$ is a finite $\mathcal{O}_S$-module).

Why affine and proper implies finite: Let $X = \text{Spec } A$ and $S = \text{Spec } B$ with $f$ corresponding to $B \to A$.

Suppose $A$ is not a finite $B$-module. Then there exist elements $a_1, a_2, \ldots \in A$ that are linearly independent over $B$.

Using these, we can construct a morphism $\text{Spec } K \to X$ (for suitable DVR $R$ with field of fractions $K$) that has no extension to $\text{Spec } R \to X$, contradicting properness.

The key idea: affine schemes have "no room to compactify," so properness forces finiteness.

Example: The morphism $\mathbb{A}^1 \to \text{Spec } k$ is not proper (even though both are affine) because $\mathbb{A}^1$ is not finite over $k$ (the coordinate ring $k[x]$ is not a finite-dimensional $k$-vector space).

Let $(A, \mathfrak{m})$ be a complete local noetherian ring (e.g., $k[[t]]$ or $\mathbb{Z}_p$).

Theorem: If $X$ is proper over $\text{Spec } A$, then the set of sections $\text{Spec } A \to X$ is in bijection with the set of sections $\text{Spec } A/\mathfrak{m} \to X$.

Proof idea:

• Given a section $\text{Spec } A/\mathfrak{m} \to X$, we can lift it inductively to $\text{Spec } A/\mathfrak{m}^n$ using properness and formal smoothness arguments.
• The completion gives a section $\text{Spec } A \to X$.

Example: Let $X = \mathbb{P}^1_k$ and $A = k[[t]]$. A $k$-rational point of $\mathbb{P}^1$ (i.e., a point in $\mathbb{P}^1(k)$) lifts uniquely to a $k[[t]]$-valued point (i.e., a point in $\mathbb{P}^1(k[[t]])$).

This is not true for $\mathbb{A}^1$: a point in $\mathbb{A}^1(k)$ doesn't uniquely lift to $\mathbb{A}^1(k[[t]])$ (there are many lifts differing by higher-order terms).

Let $C$ be a smooth curve over a field $k$.

Fact: $C$ is proper over $k$ if and only if $C$ is complete (i.e., every rational function on $C$ is constant, or equivalently, $C$ has no "missing points").

Example 1: $\mathbb{P}^1$ is proper (and smooth and complete).

Example 2: $\mathbb{A}^1$ is smooth but not complete (it's missing the point at infinity).

Example 3: Let $C \subset \mathbb{A}^2$ be the affine curve $y^2 = x^3 + x$. This is smooth but not proper (it's an open subset of its projective closure).

The projective closure $\overline{C} \subset \mathbb{P}^2$ is given by the homogeneous equation $Y^2 Z = X^3 + XZ^2$. This is a proper (in fact, projective) curve.

The valuative criterion tells us that $C$ is not proper: there are morphisms $\text{Spec } k((t)) \to C$ that don't extend to $\text{Spec } k[[t]] \to C$ (they "escape to the point at infinity").

Consider the affine line with doubled origin from Example 4.

This morphism $X \to \text{Spec } k$ is:

• Universally closed: Every morphism $\text{Spec } K \to X$ extends to $\text{Spec } R \to X$ (in fact, it has two extensions)
• Not separated: The extension is not unique

Therefore, $X$ is universally closed but not proper.

This shows that universal closedness and separatedness are independent conditions, and properness requires both.

Question: Is the morphism $f: \text{Spec } k[x,y]/(xy) \to \text{Spec } k$ universally closed?

The scheme $\text{Spec } k[x,y]/(xy)$ is the union of two lines meeting at a point (the coordinate axes in $\mathbb{A}^2$).

Answer: No, it's not universally closed.

Proof: Consider the DVR $R = k[[t]]$ and the morphism $\text{Spec } k((t)) \to \text{Spec } k[x,y]/(xy)$ given by $x = t^{-1}$, $y = t$.

We have $xy = t^{-1} \cdot t = 1 \neq 0$, so this doesn't map into $\text{Spec } k[x,y]/(xy)$.

Wait, let me reconsider. We need $xy = 0$, so if $x = t^{-1}$ and $y = t$, then $xy = 1 \neq 0$.

Let's try: $x = t$, $y = 0$. Then $xy = 0$, and this is a morphism $\text{Spec } k((t)) \to \text{Spec } k[x,y]/(xy)$.

This extends to $\text{Spec } k[[t]] \to \text{Spec } k[x,y]/(xy)$ by $x = t$, $y = 0$.

So actually, this morphism IS universally closed (it's even proper, since it's the affine line).

Better example: Let me reconsider the question. Actually, $\text{Spec } k[x,y]/(xy) \to \text{Spec } k$ IS proper because it's a projective morphism (it can be realized as a closed subscheme of $\mathbb{P}^1 \times \mathbb{P}^1$).

One of the most important applications of the valuative criterion is to moduli problems, particularly in the theory of stable curves.

Setup: Let $\mathcal{M}_g$ be the moduli space of smooth curves of genus $g \geq 2$ over a field $k$. This is not a complete variety (it's missing the boundary).

To compactify $\mathcal{M}_g$, we need to add "degenerate curves" (nodal curves). The resulting space $\overline{\mathcal{M}}_g$ (the moduli space of stable curves) is proper over $k$.

Valuative criterion interpretation: Let $R$ be a DVR with field of fractions $K$, and let $\text{Spec } K \to \mathcal{M}_g$ be a morphism (i.e., a smooth curve $C_K$ of genus $g$ over $K$).

The valuative criterion for properness of $\overline{\mathcal{M}}_g$ says:

• There exists a unique extension $\text{Spec } R \to \overline{\mathcal{M}}_g$
• This corresponds to a stable reduction: a proper flat morphism $\mathcal{C} \to \text{Spec } R$ whose generic fiber is $C_K$ and whose special fiber is a stable curve

Example: Let $R = k[[t]]$ and consider the family of elliptic curves: $y^2 = x(x-1)(x-t)$ over $\text{Spec } k((t))$.

As $t \to 0$, this degenerates to $y^2 = x^2(x-1)$ (a nodal cubic curve).

The stable reduction gives a model of this family over $\text{Spec } k[[t]]$ where the special fiber is the nodal curve (which is stable).

This is a concrete instance of the valuative criterion: the smooth curve over $k((t))$ extends uniquely to a stable curve over $k[[t]]$.

More generally, the valuative criterion is used to prove properness of moduli spaces.

Theorem (Properness of moduli spaces): Let $\mathcal{M}$ be a moduli space (or more generally, a moduli stack) representing a functor $F: \text{Schemes}/k \to \text{Sets}$. Then $\mathcal{M}$ is proper over $k$ if and only if:

1. Separatedness: Given a DVR $R$ and two objects $\xi_1, \xi_2 \in F(\text{Spec } R)$ that agree over $\text{Spec } K$, we have $\xi_1 = \xi_2$.

2. Universal closedness: Given a DVR $R$ and an object $\eta \in F(\text{Spec } K)$, there exists an extension $\xi \in F(\text{Spec } R)$ with $\xi|_K = \eta$.

This is exactly the valuative criterion, phrased in terms of the moduli functor.

Example: For the moduli space of stable curves $\overline{\mathcal{M}}_g$:

• Separatedness says: two stable curves over $R$ that are isomorphic over $K$ are isomorphic over $R$
• Universal closedness says: every smooth curve over $K$ extends to a stable curve over $R$ (stable reduction theorem)

The valuative criterion is closely related to specialization maps in algebraic geometry.

Setup: Let $f: X \to S$ be proper, and let $s \to S$ be a morphism with $s = \text{Spec } \kappa(s)$ (a geometric point).

For any point $x \in X_s$ (the fiber over $s$), we can "lift" $x$ to the generic point using the valuative criterion.

Theorem: If $f: X \to S$ is proper and $S = \text{Spec } R$ for a DVR $R$, then the specialization map $X_K \to X_\kappa$ (from the generic fiber to the special fiber) is well-defined and surjective.

Proof sketch:

• Given a point $x_\kappa \in X_\kappa$, we get a morphism $\text{Spec } \kappa \to X_\kappa \subset X$.
• Compose with $\text{Spec } \kappa \to \text{Spec } R$ to get a diagram:

• Since $\kappa$ is a field, we can find a DVR $R' \supset R$ with residue field $\kappa$ and field of fractions $K'$.
• By the valuative criterion, the morphism $\text{Spec } K' \to X$ extends to $\text{Spec } R' \to X$, giving a point in $X_K$.

This is used extensively in the study of degenerations and limit linear series.

A Fano variety is a smooth projective variety $X$ with ample anticanonical bundle $-K_X$.

Fact: All Fano varieties are proper (in fact, projective).

The valuative criterion provides a powerful tool for studying degenerations of Fano varieties:

Question: Given a family of Fano varieties over $\text{Spec } k((t))$, does it extend to a family over $\text{Spec } k[[t]]$?

Answer: Not always! The central fiber may not be smooth, and the anticanonical bundle may not be ample.

However, there is a notion of K-stability that ensures the existence of a "stable limit":

Theorem (Stable reduction for Fano varieties): A family of Fano varieties over $\text{Spec } k((t))$ extends to a family of K-stable Fano varieties over $\text{Spec } k[[t]]$ if and only if the generic fiber is K-polystable.

This is a recent deep result (Odaka, Spotti, Sun, etc.) that uses the valuative criterion extensively.