The Stacks project

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. $\square$

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. $\square$

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). $\square$

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. $\square$

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. $\square$

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. $\square$

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. $\square$

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 proper¹. By Cohomology of Schemes, Lemma 30.21.2² we conclude that $f$ is finite. Hence $U = f^{-1}(D_{+}(T_0))$ is affine. $\square$

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. $\square$