Definition 33.43.1. Let $k$ be a field. A *curve* is a variety of dimension $1$ over $k$.

## 33.43 Curves

In the Stacks project we will use the following as our definition of a curve.

Two standard examples of curves over $k$ are the affine line $\mathbf{A}^1_ k$ and the projective line $\mathbf{P}^1_ k$. The scheme $X = \mathop{\mathrm{Spec}}(k[x, y]/(f))$ is a curve if and only if $f \in k[x, y]$ is irreducible.

Our definition of a curve has the same problems as our definition of a variety, see the discussion following Definition 33.3.1. Moreover, it means that every curve comes with a specified field of definition. For example $X = \mathop{\mathrm{Spec}}(\mathbf{C}[x])$ is a curve over $\mathbf{C}$ but we can also view it as a curve over $\mathbf{R}$. The scheme $\mathop{\mathrm{Spec}}(\mathbf{Z})$ isn't a curve, even though the schemes $\mathop{\mathrm{Spec}}(\mathbf{Z})$ and $\mathbf{A}^1_{\mathbf{F}_ p}$ behave similarly in many respects.

Lemma 33.43.2. Let $X$ be a separated, irreducible scheme of dimension $> 0$ over a field $k$. Let $x \in X$ be a closed point. The open subscheme $X \setminus \{ x\} $ is not proper over $k$.

**Proof.**
Since $X$ is irreducible, $U = X \setminus \{ x\} $ is not closed in $X$. In particular, the immersion $U \to X$ is not proper. By Morphisms, Lemma 29.41.7 (here we use $X$ is separated), $U \to \mathop{\mathrm{Spec}}(k)$ is not proper either.
$\square$

Lemma 33.43.3. Let $X$ be a separated finite type scheme over a field $k$. If $\dim (X) \leq 1$ then $X$ is H-quasi-projective over $k$.

**Proof.**
By Proposition 33.38.12 the scheme $X$ has an ample invertible sheaf $\mathcal{L}$. By Morphisms, Lemma 29.39.3 we see that $X$ is isomorphic to a locally closed subscheme of $\mathbf{P}^ n_ k$ over $\mathop{\mathrm{Spec}}(k)$. This is the definition of being H-quasi-projective over $k$, see Morphisms, Definition 29.40.1.
$\square$

Lemma 33.43.4. Let $X$ be a proper scheme over a field $k$. If $\dim (X) \leq 1$ then $X$ is H-projective over $k$.

**Proof.**
By Lemma 33.43.3 we see that $X$ is a locally closed subscheme of $\mathbf{P}^ n_ k$ for some field $k$. Since $X$ is proper over $k$ it follows that $X$ is a closed subscheme of $\mathbf{P}^ n_ k$ (Morphisms, Lemma 29.41.7).
$\square$

Lemma 33.43.5. Let $X$ be a separated scheme of finite type over $k$. If $\dim (X) \leq 1$, then there exists an open immersion $j : X \to \overline{X}$ with the following properties

$\overline{X}$ is H-projective over $k$, i.e., $\overline{X}$ is a closed subscheme of $\mathbf{P}^ d_ k$ for some $d$,

$j(X) \subset \overline{X}$ is dense and scheme theoretically dense,

$\overline{X} \setminus X = \{ x_1, \ldots , x_ n\} $ for some closed points $x_ i \in \overline{X}$.

**Proof.**
By Lemma 33.43.3 we may assume $X$ is a locally closed subscheme of $\mathbf{P}^ d_ k$ for some $d$. Let $\overline{X} \subset \mathbf{P}^ d_ k$ be the scheme theoretic image of $X \to \mathbf{P}^ d_ k$, see Morphisms, Definition 29.6.2. The description in Morphisms, Lemma 29.7.7 gives properties (1) and (2). Then $\dim (X) = 1 \Rightarrow \dim (\overline{X}) = 1$ for example by looking at generic points, see Lemma 33.20.3. As $\overline{X}$ is Noetherian, it then follows that $\overline{X} \setminus X = \{ x_1, \ldots , x_ n\} $ is a finite set of closed points.
$\square$

Lemma 33.43.6. Let $X$ be a separated scheme of finite type over $k$. If $X$ is reduced and $\dim (X) \leq 1$, then there exists an open immersion $j : X \to \overline{X}$ such that

$\overline{X}$ is H-projective over $k$, i.e., $\overline{X}$ is a closed subscheme of $\mathbf{P}^ d_ k$ for some $d$,

$j(X) \subset \overline{X}$ is dense and scheme theoretically dense,

$\overline{X} \setminus X = \{ x_1, \ldots , x_ n\} $ for some closed points $x_ i \in \overline{X}$,

the local rings $\mathcal{O}_{\overline{X}, x_ i}$ are discrete valuation rings for $i = 1, \ldots , n$.

**Proof.**
Let $j : X \to \overline{X}$ be as in Lemma 33.43.5. Consider the normalization $X'$ of $\overline{X}$ in $X$. By Lemma 33.27.3 the morphism $X' \to \overline{X}$ is finite. By Morphisms, Lemma 29.44.16 $X' \to \overline{X}$ is projective. By Morphisms, Lemma 29.43.16 we see that $X' \to \overline{X}$ is H-projective. By Morphisms, Lemma 29.43.7 we see that $X' \to \mathop{\mathrm{Spec}}(k)$ is H-projective. Let $\{ x'_1, \ldots , x'_ m\} \subset X'$ be the inverse image of $\{ x_1, \ldots , x_ n\} = \overline{X} \setminus X$. Then $\dim (\mathcal{O}_{X', x'_ i}) = 1$ for all $1 \leq i \leq m$. Hence the local rings $\mathcal{O}_{X', x'}$ are discrete valuation rings by Morphisms, Lemma 29.53.16. Then $X \to X'$ and $\{ x'_1, \ldots , x'_ m\} $ is as desired.
$\square$

Lemma 33.43.7. Let $X$ be a separated scheme of finite type over $k$ with $\dim (X) \leq 1$. Then there exists a commutative diagram

of schemes with the following properties:

$k'/k$ is a finite purely inseparable extension of fields,

$\nu $ is the normalization of $X_{k'}$,

$j$ is an open immersion with dense image,

$k'_ i/k'$ is a finite separable extension for $i = 1, \ldots , n$,

$\overline{Y}_ i$ is smooth, projective, geometrically irreducible dimension $\leq 1$ over $k'_ i$.

**Proof.**
As we may replace $X$ by its reduction, we may and do assume $X$ is reduced. Choose $X \to \overline{X}$ as in Lemma 33.43.6. If we can show the lemma for $\overline{X}$, then the lemma follows for $X$ (details omitted). Thus we may and do assume $X$ is projective.

Choose $k'/k$ finite purely inseparable such that the normalization of $X_{k'}$ is geometrically normal over $k'$, see Lemma 33.27.4. Denote $Y = (X_{k'})^\nu $ the normalization; for properties of the normalization, see Section 33.27. Then $Y$ is geometrically regular as normal and regular are the same in dimension $\leq 1$, see Properties, Lemma 28.12.6. Hence $Y$ is smooth over $k'$ by Lemma 33.12.6. Let $Y = Y_1 \amalg \ldots \amalg Y_ n$ be the decomposition of $Y$ into irreducible components. Set $k'_ i = \Gamma (Y_ i, \mathcal{O}_{Y_ i})$. These are finite separable extensions of $k'$ by Lemma 33.9.3. The proof is finished by Lemma 33.9.4. $\square$

Lemma 33.43.8. Let $k$ be a field. Let $X$ be a curve over $k$. Let $x \in X$ be a closed point. We think of $x$ as a (reduced) closed subscheme of $X$ with sheaf of ideals $\mathcal{I}$. The following are equivalent

$\mathcal{O}_{X, x}$ is regular,

$\mathcal{O}_{X, x}$ is normal,

$\mathcal{O}_{X, x}$ is a discrete valuation ring,

$\mathcal{I}$ is an invertible $\mathcal{O}_ X$-module,

$x$ is an effective Cartier divisor on $X$.

If $k$ is perfect or if $\kappa (x)$ is separable over $k$, these are also equivalent to

$X \to \mathop{\mathrm{Spec}}(k)$ is smooth at $x$.

**Proof.**
Since $X$ is a curve, the local ring $\mathcal{O}_{X, x}$ is a Noetherian local domain of dimension $1$ (Lemma 33.20.3). Parts (4) and (5) are equivalent by definition and are equivalent to $\mathcal{I}_ x = \mathfrak m_ x \subset \mathcal{O}_{X, x}$ having one generator (Divisors, Lemma 31.15.2). The equivalence of (1), (2), (3), (4), and (5) therefore follows from Algebra, Lemma 10.119.7. The final statement follows from Lemma 33.25.8 in case $k$ is perfect. If $\kappa (x)/k$ is separable, then the equivalence follows from Algebra, Lemma 10.140.5.
$\square$

Remark 33.43.9. Let $k$ be a field. Let $X$ be a regular curve over $k$. By Lemmas 33.43.8 and 33.43.6 there exists a nonsingular projective curve $\overline{X}$ which is a compactification of $X$, i.e., there exists an open immersion $j : X \to \overline{X}$ such that the complement consists of a finite number of closed points. If $k$ is perfect, then $X$ and $\overline{X}$ are smooth over $k$ and $\overline{X}$ is a smooth projective compactification of $X$.

Observe that if an affine scheme $X$ over $k$ is proper over $k$ then $X$ is finite over $k$ (Morphisms, Lemma 29.44.11) and hence has dimension $0$ (Algebra, Lemma 10.53.2 and Proposition 10.60.7). Hence a scheme of dimension $> 0$ over $k$ cannot be both affine and proper over $k$. Thus the possibilities in the following lemma are mutually exclusive.

Lemma 33.43.10. Let $X$ be a curve over $k$. Then either $X$ is an affine scheme or $X$ is H-projective over $k$.

**Proof.**
Choose $X \to \overline{X}$ with $\overline{X} \setminus X = \{ x_1, \ldots , x_ r\} $ as in Lemma 33.43.6. Then $\overline{X}$ is a curve as well. If $r = 0$, then $X = \overline{X}$ is H-projective over $k$. Thus we may assume $r \geq 1$ and our goal is to show that $X$ is affine. By Lemma 33.38.2 it suffices to show that $\overline{X} \setminus \{ x_1\} $ is affine. This reduces us to the claim stated in the next paragraph.

Let $X$ be an H-projective curve over $k$. Let $x \in X$ be a closed point such that $\mathcal{O}_{X, x}$ is a discrete valuation ring. Claim: $U = X \setminus \{ x\} $ is affine. By Lemma 33.43.8 the point $x$ defines an effective Cartier divisor of $X$. For $n \geq 1$ denote $nx = x + \ldots + x$ the $n$-fold sum, see Divisors, Definition 31.13.6. Denote $\mathcal{O}_{nx}$ the structure sheaf of $nx$ viewed as a coherent module on $X$. Since every invertible module on the local scheme $nx$ is trivial the first short exact sequence of Divisors, Remark 31.14.11 reads

in our case. Note that $\dim _ k H^0(X, \mathcal{O}_{nx}) \geq n$. Namely, by Lemma 33.33.3 we have $H^0(X, \mathcal{O}_{nx}) = \mathcal{O}_{X, x}/(\pi ^ n)$ where $\pi $ in $\mathcal{O}_{X, x}$ is a uniformizer and the powers $\pi ^ i$ map to $k$-linearly independent elements in $\mathcal{O}_{X, x}/(\pi ^ n)$ for $i = 0, 1, \ldots , n - 1$. We have $\dim _ k H^1(X, \mathcal{O}_ X) < \infty $ by Cohomology of Schemes, Lemma 30.19.2. If $n > \dim _ k H^1(X, \mathcal{O}_ X)$ we conclude from the long exact cohomology sequence that there exists an $s \in \Gamma (X, \mathcal{O}_ X(nx))$ which is not a section of $\mathcal{O}_ X$. If we take $n$ minimal with this property, then $s$ will map to a generator of the stalk $\left(\mathcal{O}_ X(nx)\right)_ x$ since otherwise it would define a section of $\mathcal{O}_ X((n - 1)x) \subset \mathcal{O}_ X(nx)$. For this $n$ we conclude that $s_0 = 1$ and $s_1 = s$ generate the invertible module $\mathcal{L} = \mathcal{O}_ X(nx)$.

Consider the corresponding morphism $f = \varphi _{\mathcal{L}, (s_0, s_1)} : X \to \mathbf{P}^1_ k$ of Constructions, Section 27.13. Observe that the inverse image of $D_{+}(T_0)$ is $U = X \setminus \{ x\} $ as the section $s_0$ of $\mathcal{L}$ only vanishes at $x$. In particular, $f$ is non-constant, i.e., $\mathop{\mathrm{Im}}(f)$ has more than one point. Hence $f$ must map the generic point $\eta $ of $X$ to the generic point of $\mathbf{P}^1_ k$. Hence if $y \in \mathbf{P}^1_ k$ is a closed point, then $f^{-1}(\{ y\} )$ is a closed set of $X$ not containing $\eta $, hence finite. Finally, $f$ is proper^{1}. By Cohomology of Schemes, Lemma 30.21.2^{2} we conclude that $f$ is finite. Hence $U = f^{-1}(D_{+}(T_0))$ is affine.
$\square$

The following lemma combined with Lemma 33.43.2 tells us that given a separated scheme $X$ of dimension $1$ and of finite type over $k$, then $X \setminus Z$ is affine, whenever the closed subset $Z$ meets every irreducible component of $X$.

Lemma 33.43.11. Let $X$ be a separated scheme of finite type over $k$. If $\dim (X) \leq 1$ and no irreducible component of $X$ is proper of dimension $1$, then $X$ is affine.

**Proof.**
Let $X = \bigcup X_ i$ be the decomposition of $X$ into irreducible components. We think of $X_ i$ as an integral scheme (using the reduced induced scheme structure, see Schemes, Definition 26.12.5). In particular $X_ i$ is a singleton (hence affine) or a curve hence affine by Lemma 33.43.10. Then $\coprod X_ i \to X$ is finite surjective and $\coprod X_ i$ is affine. Thus we see that $X$ is affine by Cohomology of Schemes, Lemma 30.13.3.
$\square$

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