Lemma 108.62.1. There exists a field $k$ and a family of curves $X \to \mathbf{A}^1_ k$ such that $X$ is not a scheme.

## 108.62 A family of curves whose total space is not a scheme

In Quot, Section 97.15 we define a family of curves over a scheme $S$ to be a proper, flat, finitely presented morphism of relative dimension $\leq 1$ from an algebraic space $X$ to $S$. If $S$ is the spectrum of a complete Noetherian local ring, then $X$ is a scheme, see More on Morphisms of Spaces, Lemma 74.43.5. In this section we show this is not true in general.

Let $k$ be a field. We start with a proper flat morphism

and a point $y \in Y(k)$ lying over $0 \in \mathbf{A}^1_ k(k)$ with the following properties

the fibre $Y_0$ is a smooth geometrically irreducible curve over $k$,

for any proper closed subscheme $T \subset Y$ dominating $\mathbf{A}^1_ k$ the intersection $T \cap Y_0$ contains at least one point distinct from $y$.

Given such a surface we construct our example as follows.

Here $Z \to Y$ is the blowup of $Y$ in $y$. Let $E \subset Z$ be the exceptional divisor and let $C \subset Z$ be the strict transform of $Y_0$. We have $Z_0 = E \cup C$ scheme theoretically (to see this use that $Y$ is smooth at $y$ and moreover $Y \to \mathbf{A}^1_ k$ is smooth at $y$). By Artin's results ([ArtinII]; use Semistable Reduction, Lemma 55.9.7 to see that the normal bundle of $C$ is negative) we can blow down the curve $C$ in $Z$ to obtain an algebraic space $X$ as in the diagram. Let $x \in X(k)$ be the image of $C$.

We claim that $X$ is not a scheme. Namely, if it were a scheme, then there would be an affine open neighbourhood $U \subset X$ of $x$. Set $T = X \setminus U$. Then $T$ dominates $\mathbf{A}^1_ k$ (as the fibres of $X \to \mathbf{A}^1_ k$ are proper of dimension $1$ and the fibres of $U \to \mathbf{A}^1_ k$ are affine hence different). Let $T' \subset Z$ be the closed subscheme mapping isomorphically to $T$ (as $x \not\in T$). Then the image of $T'$ in $X$ contradicts condition (2) above (as $T' \cap Z_0$ is contained in the exceptional divisor $E$ of the blowing up $Z \to Y$).

To finish the discussion we need to construct our $Y$. We will assume the characteristic of $k$ is not $3$. Write $\mathbf{A}^1_ k = \mathop{\mathrm{Spec}}(k[t])$ and take

in $\mathbf{P}^2_{k[t]}$. The fibre of this for $t = 0$ is a smooth projective genus $1$ curve. On the affine piece $V_+(T_0)$ we get the affine equation

which defines a smooth surface over $k$. Since the same is true on the other affine pieces by symmetry we see that $Y$ is a smooth surface. Finally, we see from the affine equation also that the fraction field is $k(x, y)$ hence $Y$ is a rational surface. Now the Picard group of a rational surface is finitely generated (insert future reference here). Hence in order to choose $y \in Y_0(k)$ with property (2) it suffices to choose $y$ such that

Namely, the sum of the $1$-dimensional irreducible components of a $T$ contradicting (2) would give an effective Cartier divisor intersection $Y_0$ in the divisor $ny$ for some $n \geq 1$ and we would conclude that $\mathcal{O}_{Y_0}(ny)$ is in the image of the restriction map. Observe that since $Y_0$ has genus $\geq 1$ the map

is injective. Now if $k$ is an uncountable algebraically closed field, then using the countability of $\mathop{\mathrm{Pic}}\nolimits (Y)$ and the remark just made, we can find a $y \in Y_0(k)$ satisfying (108.62.0.1) and hence (2).

**Proof.**
See discussion above.
$\square$

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