Proposition 53.10.4 (Characterization of the projective line). Let $k$ be a field. Let $X$ be a proper curve over $k$. The following are equivalent

1. $X \cong \mathbf{P}^1_ k$,

2. $X$ is smooth and geometrically irreducible over $k$, $X$ has genus $0$, and $X$ has an invertible module of odd degree,

3. $X$ is geometrically integral over $k$, $X$ has genus $0$, $X$ is Gorenstein, and $X$ has an invertible sheaf of odd degree,

4. $H^0(X, \mathcal{O}_ X) = k$, $X$ has genus $0$, $X$ is Gorenstein, and $X$ has an invertible sheaf of odd degree,

5. $X$ is geometrically integral over $k$, $X$ has genus $0$, and $X$ has an invertible $\mathcal{O}_ X$-module of degree $1$,

6. $H^0(X, \mathcal{O}_ X) = k$, $X$ has genus $0$, and $X$ has an invertible $\mathcal{O}_ X$-module of degree $1$,

7. $H^1(X, \mathcal{O}_ X) = 0$ and $X$ has an invertible $\mathcal{O}_ X$-module of degree $1$,

8. $H^1(X, \mathcal{O}_ X) = 0$ and $X$ has closed points $x_1, \ldots , x_ n$ such that $\mathcal{O}_{X, x_ i}$ is normal and $\gcd ([\kappa (x_ i) : k]) = 1$, and

9. add more here.

Proof. We will prove that each condition (2) – (8) implies (1) and we omit the verification that (1) implies (2) – (8).

Assume (2). A smooth scheme over $k$ is geometrically reduced (Varieties, Lemma 33.25.4) and regular (Varieties, Lemma 33.25.3). Hence $X$ is Gorenstein (Duality for Schemes, Lemma 48.24.3). Thus we reduce to (3).

Assume (3). Since $X$ is geometrically integral over $k$ we have $H^0(X, \mathcal{O}_ X) = k$ by Varieties, Lemma 33.26.2. and we reduce to (4).

Assume (4). Since $X$ is Gorenstein the dualizing module $\omega _ X$ as in Lemma 53.4.1 has degree $\deg (\omega _ X) = -2$ by Lemma 53.8.3. Combined with the assumed existence of an odd degree invertible module, we conclude there exists an invertible module of degree $1$. In this way we reduce to (6).

Assume (5). Since $X$ is geometrically integral over $k$ we have $H^0(X, \mathcal{O}_ X) = k$ by Varieties, Lemma 33.26.2. and we reduce to (6).

Assume (6). Then $X \cong \mathbf{P}^1_ k$ by Lemma 53.10.2.

Assume (7). Observe that $\kappa = H^0(X, \mathcal{O}_ X)$ is a field finite over $k$ by Varieties, Lemma 33.26.2. If $d = [\kappa : k] > 1$, then every invertible sheaf has degree divisible by $d$ and there cannot be an invertible sheaf of degree $1$. Hence $d = 1$ and we reduce to case (6).

Assume (8). Observe that $\kappa = H^0(X, \mathcal{O}_ X)$ is a field finite over $k$ by Varieties, Lemma 33.26.2. Since $\kappa \subset \kappa (x_ i)$ we see that $k = \kappa$ by the assumption on the gcd of the degrees. The same condition allows us to find integers $a_ i$ such that $1 = \sum a_ i[\kappa (x_ i) : k]$. Because $x_ i$ defines an effective Cartier divisor on $X$ by Varieties, Lemma 33.43.16 we can consider the invertible module $\mathcal{O}_ X(\sum a_ i x_ i)$. By our choice of $a_ i$ the degree of $\mathcal{L}$ is $1$. Thus $X \cong \mathbf{P}^1_ k$ by Lemma 53.10.2. $\square$

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