## 53.10 Curves of genus zero

Later we will need to know what a proper genus zero curve looks like. It turns out that a Gorenstein proper genus zero curve is a plane curve of degree $2$, i.e., a conic, see Lemma 53.10.3. A general proper genus zero curve is obtained from a nonsingular one (over a bigger field) by a pushout procedure, see Lemma 53.10.5. Since a nonsingular curve is Gorenstein, these two results cover all possible cases.

Lemma 53.10.1. Let $X$ be a proper curve over a field $k$ with $H^0(X, \mathcal{O}_ X) = k$. If $X$ has genus $0$, then every invertible $\mathcal{O}_ X$-module $\mathcal{L}$ of degree $0$ is trivial.

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 33.43.12. $\square$

Lemma 53.10.2. Let $X$ be a proper curve over a field $k$ with $H^0(X, \mathcal{O}_ X) = k$. Assume $X$ has genus $0$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module of degree $d > 0$. Then we have

1. $\dim _ k H^0(X, \mathcal{L}) = d + 1$ and $\dim _ k H^1(X, \mathcal{L}) = 0$,

2. $\mathcal{L}$ is very ample and defines a closed immersion into $\mathbf{P}^ d_ k$.

Proof. By definition of degree and genus we have

$\dim _ k H^0(X, \mathcal{L}) - \dim _ k H^1(X, \mathcal{L}) = d + 1$

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

$0 \to \mathcal{O}_ X \to \mathcal{L} \to \mathcal{L}|_ D \to 0$

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 33.32.3) 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

$\mu _ n : H^0(X, \mathcal{L}) \otimes _ k H^0(X, \mathcal{L}^{\otimes n}) \longrightarrow H^0(X, \mathcal{L}^{\otimes n + 1})$

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

$0 \to \mathcal{L}^{\otimes n} \xrightarrow {s} \mathcal{L}^{\otimes n + 1} \to \mathcal{L}^{\otimes n + 1}|_ D \to 0$

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 33.43.14. Hence Morphisms, Lemma 29.43.17 gives an isomorphism

$X \longrightarrow \text{Proj}\left( \bigoplus \nolimits _{n \geq 0} H^0(X, \mathcal{L}^{\otimes n})\right)$

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 $\mathcal{L}$. $\square$

Lemma 53.10.3. Let $X$ be a proper curve over a field $k$ with $H^0(X, \mathcal{O}_ X) = k$. If $X$ is Gorenstein and has genus $0$, then $X$ is isomorphic to a plane curve of degree $2$.

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

$\varphi _{(\mathcal{L}, (s_0, s_1, s_2))} : X \longrightarrow \mathbf{P}^2_ k$

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. $\square$

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. $\square$

Lemma 53.10.5. Let $X$ be a proper curve over a field $k$ with $H^0(X, \mathcal{O}_ X) = k$. Assume $X$ is singular and has genus $0$. Then there exists a diagram

$\xymatrix{ x' \ar[d] \ar[r] & X' \ar[d]^\nu \ar[r] & \mathop{\mathrm{Spec}}(k') \ar[d] \\ x \ar[r] & X \ar[r] & \mathop{\mathrm{Spec}}(k) }$

where

1. $k'/k$ is a nontrivial finite extension,

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

3. $x'$ is a $k'$-rational point of $X'$,

4. $x$ is a $k$-rational point of $X$,

5. $X' \setminus \{ x'\} \to X \setminus \{ x\}$ is an isomorphism,

6. $0 \to \mathcal{O}_ X \to \nu _*\mathcal{O}_{X'} \to k'/k \to 0$ is a short exact sequence where $k'/k = \kappa (x')/\kappa (x)$ indicates the skyscraper sheaf on the point $x$.

Proof. Let $\nu : X' \to X$ be the normalization of $X$, see Varieties, Sections 33.27 and 33.40. 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

$0 \to \mathcal{O}_ X \to \nu _*\mathcal{O}_{X'} \to \mathcal{Q} \to 0$

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

$\mathcal{I} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\nu _*\mathcal{O}_{X'}, \mathcal{O}_ X)$

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

$0 \to \mathcal{I} \to \mathcal{O}_ X \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{Q}, \mathcal{Q})$

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

$0 \to \mathcal{I} \to \mathcal{O}_ X \to \mathcal{A} \to 0$

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:

$0 \to \mathcal{I}' \to \mathcal{O}_{X'} \to \mathcal{A}' \to 0$

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

$\dim _ k H^0(X', \mathcal{A}') = \dim _ k H^0(X, \mathcal{A}) + \dim _ k H^0(X, \mathcal{Q}) \leq 2 n - 2$

However, since $X'$ is a curve over $k'$ we see that the left hand side is divisible by $n$ (Varieties, Lemma 33.43.10). 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. $\square$

Example 53.10.6. In fact, the situation described in Lemma 53.10.5 occurs for any nontrivial finite extension $k'/k$. Namely, we can consider

$A = \{ f \in k'[x] \mid f(0) \in k \}$

The spectrum of $A$ is an affine curve, which we can glue to the spectrum of $B = k'[y]$ using the isomorphism $A_ x \cong B_ y$ sending $x^{-1}$ to $y$. The result is a proper curve $X$ with $H^0(X, \mathcal{O}_ X) = k$ and singular point $x$ corresponding to the maximal ideal $A \cap (x)$. The normalization of $X$ is $\mathbf{P}^1_{k'}$ exactly as in the lemma.

[1] In our case this follows from Divisors, Lemma 31.17.1 as $D \to \mathop{\mathrm{Spec}}(k)$ is finite.

Comment #2999 by Ariyan on

Typo in first paragraph: "genus zero curve is obtain from " -> "genus zero curve is obtained from"

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).