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The Stacks project

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 33.43.8 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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