Consider $f: \mathbb{A}^2_k \to \operatorname{Spec} k$ over a field $k$. Then $\mathbb{A}^2_k \times_k \mathbb{A}^2_k \cong \mathbb{A}^4_k$ with coordinates $(x_1, y_1, x_2, y_2)$.

The diagonal morphism $\Delta: \mathbb{A}^2_k \to \mathbb{A}^4_k$ is given by: $(x, y) \mapsto (x, y, x, y)$

The image of $\Delta$ is the closed subscheme of $\mathbb{A}^4_k$ defined by the ideal $(x_1 - x_2, y_1 - y_2)$, which is isomorphic to $\mathbb{A}^2_k$.

Let $X = \operatorname{Spec} A$ and $S = \operatorname{Spec} R$ with $f: X \to S$ corresponding to a ring homomorphism $\varphi: R \to A$.

Then $X \times_S X = \operatorname{Spec}(A \otimes_R A)$, and the diagonal morphism corresponds to the multiplication map: $m: A \otimes_R A \to A, \quad a \otimes b \mapsto ab$

The kernel of $m$ is the ideal $I$ generated by elements of the form $a \otimes 1 - 1 \otimes a$ for $a \in A$. The diagonal is the closed subscheme $\operatorname{Spec}(A) \subseteq \operatorname{Spec}(A \otimes_R A)$.

For a topological space $X$, the space is Hausdorff if and only if the diagonal $\Delta = \{(x, x) : x \in X\} \subseteq X \times X$ is closed.

Proof: If $X$ is Hausdorff and $(x, y) \notin \Delta$ (so $x \neq y$), there exist disjoint open sets $U, V$ with $x \in U$ and $y \in V$. Then $U \times V$ is an open neighborhood of $(x, y)$ disjoint from $\Delta$. Thus the complement of $\Delta$ is open.

Conversely, if $\Delta$ is closed and $x \neq y$, then $(x, y) \in (X \times X) \setminus \Delta$, which is open. This open set contains a basic open $U \times V$ with $x \in U$ and $y \in V$, separating $x$ and $y$.

Any affine scheme $X = \operatorname{Spec} A$ is separated over $\operatorname{Spec} \mathbb{Z}$.

The fiber product $X \times X = \operatorname{Spec}(A \otimes_{\mathbb{Z}} A)$, and the diagonal corresponds to multiplication $A \otimes_{\mathbb{Z}} A \to A$. The kernel is generated by $a \otimes 1 - 1 \otimes a$, which cuts out a closed subscheme isomorphic to $X$.

Projective space $\mathbb{P}^n_S$ over any scheme $S$ is separated over $S$.

The key observation is that $\mathbb{P}^n_S$ can be covered by affine open subschemes $U_i \cong \mathbb{A}^n_S$, and on overlaps $U_i \cap U_j$, the transition maps are given by Laurent polynomials, which define closed immersions on appropriate affine patches.

The diagonal $\Delta: \mathbb{P}^n_S \to \mathbb{P}^n_S \times_S \mathbb{P}^n_S$ can be shown to be a closed immersion by checking on the standard affine open cover.

The canonical example of a non-separated scheme is the line with doubled origin.

Construction: Start with two copies of the affine line $\mathbb{A}^1_k = \operatorname{Spec} k[t]$. Call them $U_1 = \operatorname{Spec} k[t_1]$ and $U_2 = \operatorname{Spec} k[t_2]$.

Remove the origin from each: let $V_1 = \operatorname{Spec} k[t_1, t_1^{-1}]$ and $V_2 = \operatorname{Spec} k[t_2, t_2^{-1}]$ be the punctured lines.

Glue $U_1$ and $U_2$ by identifying $V_1 \cong V_2$ via the isomorphism $t_1 \mapsto t_2$.

The resulting scheme $X$ is the line with doubled origin. It has two points $0_1 \in U_1$ and $0_2 \in U_2$ that both "look like" the origin, but they remain distinct in $X$.

Why it's not separated: Consider the diagonal $\Delta: X \to X \times_k X$. The two points $(0_1, 0_2)$ and $(0_2, 0_1)$ in $X \times_k X$ lie in the closure of the diagonal (since points near the origins can be paired), but they are not actually on the diagonal itself (since $0_1 \neq 0_2$ in $X$).

More precisely, the diagonal is not closed because its complement contains points that are limits of diagonal points but not themselves diagonal.

Let $X$ be the line with doubled origin. Consider the fiber product $X \times_k X$.

The complement of the diagonal consists of points $(x, y)$ where $x \neq y$ in $X$. However, the points $(0_1, 0_2)$ and $(0_2, 0_1)$ are in the closure of this set (topologically), yet they represent distinct points that "should" be separated but aren't.

If we consider a morphism from $\operatorname{Spec} k[\epsilon]/(\epsilon^2)$ to $X$ sending the closed point to $0_1$ and the generic point into the common part, and another sending the closed point to $0_2$, we get a morphism to $X \times_k X$ whose image approaches the "non-diagonal" points $(0_1, 0_2)$.

This violates the Hausdorff-like property: we cannot separate the two origins.

More generally, take any scheme $Y$ and two copies of it, $U_1$ and $U_2$. Let $V \subseteq Y$ be an open subscheme, and identify $V \subseteq U_1$ with $V \subseteq U_2$.

If $V \neq Y$, the resulting scheme $X$ obtained by gluing $U_1$ and $U_2$ along $V$ is typically not separated, as points in $Y \setminus V$ appear with multiplicity.

In topology, a continuous map $f: X \to Y$ between locally compact Hausdorff spaces is proper if $f$ is closed and preimages of points are compact.

For schemes, proper morphisms play the same role:

• Separated corresponds to Hausdorff (diagonal is closed),
• Universally closed corresponds to compactness (preimages are "compact"),
• Finite type ensures reasonable fibers.

Just as compact Hausdorff spaces have excellent properties in topology, proper schemes have excellent properties in algebraic geometry.

Any finite morphism $f: X \to S$ is proper.

Finite morphisms are affine and of finite type, hence separated. They are also universally closed because the image of a closed set under $\operatorname{Spec} B \to \operatorname{Spec} A$ (where $B$ is a finite $A$-algebra) is closed, and this property is preserved under base change.

Any closed immersion $i: Z \to X$ is proper.

Closed immersions are separated, of finite type, and universally closed (being closed maps that remain closed under base change).

The morphism $\mathbb{P}^n_S \to S$ is proper for any scheme $S$.

We've already seen that $\mathbb{P}^n_S$ is separated over $S$. It's clearly of finite type (it has a finite cover by affine schemes, each of finite type over $S$).

The universal closedness is the deepest part: this is essentially the classical theorem that the image of a projective variety under a regular map is closed (after taking into account all base changes).

A morphism $f: X \to S$ is projective if it factors as $X \xrightarrow{i} \mathbb{P}^n_S \xrightarrow{\pi} S$ where $i$ is a closed immersion.

Since closed immersions and $\mathbb{P}^n_S \to S$ are proper, and properness is preserved under composition, every projective morphism is proper.

This is one of the most important sources of proper morphisms in algebraic geometry.

Consider $f: X \to \operatorname{Spec} k$ where $X$ is a complete variety over a field $k$. Then $f_*\mathcal{O}_X$ is a coherent sheaf on $\operatorname{Spec} k$, which means it's a finite-dimensional $k$-vector space.

In fact, $f_*\mathcal{O}_X = H^0(X, \mathcal{O}_X)$, the space of global sections. For $X$ connected, if $X$ is also reduced, then $H^0(X, \mathcal{O}_X) = k$ (the constant functions).

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

Let $R = k[[t]]$ with fraction field $K = k((t))$. Given a morphism $\operatorname{Spec} K \to X$, corresponding to a ring map $k[x] \to k((t))$, say $x \mapsto \sum_{i=n}^{\infty} a_i t^i$ with $a_n \neq 0$ and $n \in \mathbb{Z}$.

If $n \geq 0$, this extends uniquely to $k[x] \to k[[t]]$ (the same formula defines an element of $k[[t]]$).

If $n < 0$, there is no extension to $k[[t]]$ (the image of $x$ would have a pole).

The uniqueness of extension (when it exists) reflects the fact that $\mathbb{A}^1_k$ is separated over $k$.

Consider $\mathbb{P}^1_k \to \operatorname{Spec} k$. Let $R = k[[t]]$ with $K = k((t))$.

Any morphism $\operatorname{Spec} K \to \mathbb{P}^1_k$ corresponds to a point in $\mathbb{P}^1(K) = K \cup \{\infty\}$.

If the point is $[a_0 : a_1]$ with $a_0, a_1 \in K$, not both zero, we can normalize so that at least one of $a_0, a_1 \in R$ is a unit.

Case 1: $a_0 \in R^*$ (unit). Then $[a_0 : a_1] = [1 : a_1/a_0]$, and $a_1/a_0 \in R$, giving a point in the standard affine chart $\operatorname{Spec} k[x] \subseteq \mathbb{P}^1_k$ with $x = a_1/a_0 \in R$.

Case 2: $a_1 \in R^*$. Then $[a_0 : a_1] = [a_0/a_1 : 1]$, giving a point in the other affine chart with coordinate $a_0/a_1 \in R$.

In all cases, we get a unique extension $\operatorname{Spec} R \to \mathbb{P}^1_k$. This demonstrates properness.

Consider $\mathbb{A}^1_k \to \operatorname{Spec} k$ with $R = k[[t]]$.

The morphism $\operatorname{Spec} k((t)) \to \mathbb{A}^1_k$ sending $x \mapsto t^{-1}$ does not extend to $\operatorname{Spec} k[[t]] \to \mathbb{A}^1_k$, because $t^{-1} \notin k[[t]]$.

Thus $\mathbb{A}^1_k$ fails the existence part of the valuative criterion for properness (though it satisfies the uniqueness part, being separated).

This corresponds to the fact that $\mathbb{A}^1$ is not compact: sequences can "escape to infinity."

Let $C$ be a smooth projective curve over a field $k$, and let $U \subseteq C$ be a non-empty open subset (so $C \setminus U$ consists of finitely many closed points).

The morphism $U \to \operatorname{Spec} k$ is separated (being an open immersion into a separated scheme) but not proper.

To see failure of properness via the valuative criterion: let $p \in C \setminus U$ be a closed point. Choose a DVR $R$ with residue field $\kappa(p)$ and a uniformizer $t$ such that $\operatorname{Spec} R$ maps to a neighborhood of $p$ in $C$ with the generic point in $U$ and the closed point at $p$.

Then $\operatorname{Spec} K \to U$ (where $K$ is the fraction field of $R$) cannot extend to $\operatorname{Spec} R \to U$ because $p \notin U$. This violates existence in the valuative criterion.

Consider $X = \mathbb{A}^1_k \setminus \{0\} \sqcup \mathbb{A}^1_k \setminus \{0\}$ (two copies of the punctured line, glued along the identity map).

This is separated over $k$ (the diagonal is closed), but not proper. To see why, consider $R = k[[t]]$ and a morphism $\operatorname{Spec} k((t)) \to X$ sending the generic point to an element $\sum_{i=n}^{\infty} a_i t^i$ with $n < 0$ in one of the copies of $k((t))$.

This morphism doesn't extend to $\operatorname{Spec} k[[t]] \to X$ because:

• It can't extend into the first copy (the image would need to be $t^{-1}$, which doesn't converge),
• It can't extend into the second copy (same reason),
• There's no point in $X$ at "infinity" where it could land.

Thus existence fails in the valuative criterion, confirming $X$ is not proper.

Let $X$ be a complete variety over an algebraically closed field $k$. By Chow's Lemma, there exists a projective variety $\tilde{X}$ and a surjective birational morphism $\pi: \tilde{X} \to X$.

If $X$ is normal, we can take $\pi$ to be the normalization of $X$ in a projective model. This shows that studying complete varieties reduces (to some extent) to studying projective varieties.