The Stacks project

Proof. Namely, we have $\dim _ k H^0(X, \mathcal{L}) \geq 0 + 1 - 0 = 1$ by Riemann-Roch (Lemma 53.5.2), hence $\mathcal{L}$ has a nonzero section, hence $\mathcal{L} \cong \mathcal{O}_ X$ by Varieties, Lemma 33.44.12. $\square$

Proof. By definition of degree and genus we have

Let $s$ be a nonzero section of $\mathcal{L}$. Then the zero scheme of $s$ is an effective Cartier divisor $D \subset X$, we have $\mathcal{L} = \mathcal{O}_ X(D)$ and we have a short exact sequence

see Divisors, Lemma 31.14.10 and Remark 31.14.11. Since $H^1(X, \mathcal{O}_ X) = 0$ by assumption, we see that $H^0(X, \mathcal{L}) \to H^0(X, \mathcal{L}|_ D)$ is surjective. As $\mathcal{L}|_ D$ is generated by global sections (because $\dim (D) = 0$, see Varieties, Lemma 33.33.3) we conclude that the invertible module $\mathcal{L}$ is generated by global sections. In fact, since $D$ is an Artinian scheme we have $\mathcal{L}|_ D \cong \mathcal{O}_ D$¹ and hence we can find a section $t$ of $\mathcal{L}$ whose restriction of $D$ generates $\mathcal{L}|_ D$. The short exact sequence also shows that $H^1(X, \mathcal{L}) = 0$.

For $n \geq 1$ consider the multiplication map

We claim this is surjective. To see this we consider the short exact sequence

The sections of $\mathcal{L}^{\otimes n + 1}$ coming from the left in this sequence are in the image of $\mu _ n$. On the other hand, since $H^0(\mathcal{L}) \to H^0(\mathcal{L}|_ D)$ is surjective and since $t^ n$ maps to a trivialization of $\mathcal{L}^{\otimes n}|_ D$ we see that $\mu _ n(H^0(X, \mathcal{L}) \otimes t^ n)$ gives a subspace of $H^0(X, \mathcal{L}^{\otimes n + 1})$ surjecting onto the global sections of $\mathcal{L}^{\otimes n + 1}|_ D$. This proves the claim.

Observe that $\mathcal{L}$ is ample by Varieties, Lemma 33.44.14. Hence Morphisms, Lemma 29.43.17 gives an isomorphism

Since the maps $\mu _ n$ are surjective for all $n \geq 1$ we see that the graded algebra on the right hand side is a quotient of the symmetric algebra on $H^0(X, \mathcal{L})$. Choosing a $k$-basis $s_0, \ldots , s_ d$ of $H^0(X, \mathcal{L})$ we see that it is a quotient of a polynomial algebra in $d + 1$ variables. Since quotients of graded rings correspond to closed immersions of $\text{Proj}$ (Constructions, Lemma 27.11.5) we find a closed immersion $X \to \mathbf{P}^ d_ k$. We omit the verification that this morphism is the morphism of Constructions, Lemma 27.13.1 associated to the sections $s_0, \ldots , s_ d$ of $\mathcal{L}$. $\square$

Proof. Consider the invertible sheaf $\mathcal{L} = \omega _ X^{\otimes -1}$ where $\omega _ X$ is as in Lemma 53.4.1. Then $\deg (\omega _ X) = -2$ by Lemma 53.8.3 and hence $\deg (\mathcal{L}) = 2$. By Lemma 53.10.2 we conclude that choosing a basis $s_0, s_1, s_2$ of the $k$-vector space of global sections of $\mathcal{L}$ we obtain a closed immersion

Thus $X$ is a plane curve of some degree $d$. Let $F \in k[T_0, T_1, T_2]_ d$ be its equation (Lemma 53.9.1). Because the genus of $X$ is $0$ we see that $d$ is $1$ or $2$ (Lemma 53.9.3). Observe that $F$ restricts to the zero section on $\varphi (X)$ and hence $F(s_0, s_1, s_2)$ is the zero section of $\mathcal{L}^{\otimes 2}$. Because $s_0, s_1, s_2$ are linearly independent we see that $F$ cannot be linear, i.e., $d = \deg (F) \geq 2$. Thus $d = 2$ and the proof is complete. $\square$

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. $\square$

Proof. Let $\nu : X' \to X$ be the normalization of $X$, see Varieties, Sections 33.27 and 33.41. Since $X$ is singular $\nu $ is not an isomorphism. Then $k' = H^0(X', \mathcal{O}_{X'})$ is a finite extension of $k$ (Varieties, Lemma 33.26.2). The short exact sequence

and the fact that $\mathcal{Q}$ is supported in finitely many closed points give us that

1. $H^1(X', \mathcal{O}_{X'}) = 0$, i.e., $X'$ has genus $0$ as a curve over $k'$,

2. there is a short exact sequence $0 \to k \to k' \to H^0(X, \mathcal{Q}) \to 0$.

In particular $k'/k$ is a nontrivial extension.

Next, we consider what is often called the conductor ideal

This is a quasi-coherent $\mathcal{O}_ X$-module. We view $\mathcal{I}$ as an ideal in $\mathcal{O}_ X$ via the map $\varphi \mapsto \varphi (1)$. Thus $\mathcal{I}(U)$ is the set of $f \in \mathcal{O}_ X(U)$ such that $f \left(\nu _*\mathcal{O}_{X'}(U)\right) \subset \mathcal{O}_ X(U)$. In other words, the condition is that $f$ annihilates $\mathcal{Q}$. In other words, there is a defining exact sequence

Let $U \subset X$ be an affine open containing the support of $\mathcal{Q}$. Then $V = \mathcal{Q}(U) = H^0(X, \mathcal{Q})$ is a $k$-vector space of dimension $n - 1$. The image of $\mathcal{O}_ X(U) \to \mathop{\mathrm{Hom}}\nolimits _ k(V, V)$ is a commutative subalgebra, hence has dimension $\leq n - 1$ over $k$ (this is a property of commutative subalgebras of matrix algebras; details omitted). We conclude that we have a short exact sequence

where $\text{Supp}(\mathcal{A}) = \text{Supp}(\mathcal{Q})$ and $\dim _ k H^0(X, \mathcal{A}) \leq n - 1$. On the other hand, the description $\mathcal{I} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\nu _*\mathcal{O}_{X'}, \mathcal{O}_ X)$ provides $\mathcal{I}$ with a $\nu _*\mathcal{O}_{X'}$-module structure such that the inclusion map $\mathcal{I} \to \nu _*\mathcal{O}_{X'}$ is a $\nu _*\mathcal{O}_{X'}$-module map. We conclude that $\mathcal{I} = \nu _*\mathcal{I}'$ for some quasi-coherent sheaf of ideals $\mathcal{I}' \subset \mathcal{O}_{X'}$, see Morphisms, Lemma 29.11.6. Define $\mathcal{A}'$ as the cokernel:

Combining the exact sequences so far we obtain a short exact sequence $0 \to \mathcal{A} \to \nu _*\mathcal{A}' \to \mathcal{Q} \to 0$. Using the estimate above, combined with $\dim _ k H^0(X, \mathcal{Q}) = n - 1$, gives

However, since $X'$ is a curve over $k'$ we see that the left hand side is divisible by $n$ (Varieties, Lemma 33.44.10). As $\mathcal{A}$ and $\mathcal{A}'$ cannot be zero, we conclude that $\dim _ k H^0(X', \mathcal{A}') = n$ which means that $\mathcal{I}'$ is the ideal sheaf of a $k'$-rational point $x'$. By Proposition 53.10.4 we find $X' \cong \mathbf{P}^1_{k'}$. Going back to the equalities above, we conclude that $\dim _ k H^0(X, \mathcal{A}) = 1$. This means that $\mathcal{I}$ is the ideal sheaf of a $k$-rational point $x$. Then $\mathcal{A} = \kappa (x) = k$ and $\mathcal{A}' = \kappa (x') = k'$ as skyscraper sheaves. Comparing the exact sequences given above, this immediately implies the result on structure sheaves as stated in the lemma. $\square$