Definition 53.3.1. Let $k$ be a field. Let $X$ be a proper scheme of dimension $\leq 1$ over $k$. Let $d \geq 0$ and $r \geq 0$. A *linear series of degree $d$ and dimension $r$* is a pair $(\mathcal{L}, V)$ where $\mathcal{L}$ is an invertible $\mathcal{O}_ X$-module of degree $d$ (Varieties, Definition 33.44.1) and $V \subset H^0(X, \mathcal{L})$ is a $k$-subvector space of dimension $r + 1$. We will abbreviate this by saying $(\mathcal{L}, V)$ is a *$\mathfrak g^ r_ d$* on $X$.

## 53.3 Linear series

We deviate from the classical story (see Remark 53.3.6) by defining linear series in the following manner.

We will mostly use this when $X$ is a nonsingular proper curve. In fact, the definition above is just one way to generalize the classical definition of a $\mathfrak g^ r_ d$. For example, if $X$ is a proper curve, then one can generalize linear series by allowing $\mathcal{L}$ to be a torsion free coherent $\mathcal{O}_ X$-module of rank $1$. On a nonsingular curve every torsion free coherent module is locally free, so this agrees with our notion for nonsingular proper curves.

The following lemma explains the geometric meaning of linear series for proper nonsingular curves.

Lemma 53.3.2. Let $k$ be a field. Let $X$ be a nonsingular proper curve over $k$. Let $(\mathcal{L}, V)$ be a $\mathfrak g^ r_ d$ on $X$. Then there exists a morphism

of varieties over $k$ and a map $\alpha : \varphi ^*\mathcal{O}_{\mathbf{P}^ r_ k}(1) \to \mathcal{L}$ such that $\varphi ^*T_0, \ldots , \varphi ^*T_ r$ are sent to a basis of $V$ by $\alpha $.

**Proof.**
Let $s_0, \ldots , s_ r \in V$ be a $k$-basis. Since $X$ is nonsingular the image $\mathcal{L}' \subset \mathcal{L}$ of the map $s_0, \ldots , s_ r : \mathcal{O}_ X^{\oplus r} \to \mathcal{L}$ is an invertible $\mathcal{O}_ X$-module for example by Divisors, Lemma 31.11.11. Then we use Constructions, Lemma 27.13.1 to get a morphism

as in the statement of the lemma. $\square$

Lemma 53.3.3. Let $k$ be a field. Let $X$ be a proper scheme of dimension $\leq 1$ over $k$. If $X$ has a $\mathfrak g^ r_ d$, then $X$ has a $\mathfrak g^ s_ d$ for all $0 \leq s \leq r$.

**Proof.**
This is true because a vector space $V$ of dimension $r + 1$ over $k$ has a linear subspace of dimension $s + 1$ for all $0 \leq s \leq r$.
$\square$

Lemma 53.3.4. Let $k$ be a field. Let $X$ be a nonsingular proper curve over $k$. Let $(\mathcal{L}, V)$ be a $\mathfrak g^1_ d$ on $X$. Then the morphism $\varphi : X \to \mathbf{P}^1_ k$ of Lemma 53.3.2 either

is nonconstant and has degree $\leq d$, or

factors through a closed point of $\mathbf{P}^1_ k$ and in this case $H^0(X, \mathcal{O}_ X) \not= k$.

**Proof.**
By Lemma 53.3.2 we see that $\mathcal{L}' = \varphi ^*\mathcal{O}_{\mathbf{P}^1_ k}(1)$ has a nonzero map $\mathcal{L}' \to \mathcal{L}$. Hence by Varieties, Lemma 33.44.12 we see that $0 \leq \deg (\mathcal{L}') \leq d$. If $\deg (\mathcal{L}') = 0$, then the same lemma tells us $\mathcal{L}' \cong \mathcal{O}_ X$ and since we have two linearly independent sections we find we are in case (2). If $\deg (\mathcal{L}') > 0$ then $\varphi $ is nonconstant (since the pullback of an invertible module by a constant morphism is trivial). Hence

by Varieties, Lemma 33.44.11. This finishes the proof as the degree of $\mathcal{O}_{\mathbf{P}^1_ k}(1)$ is $1$. $\square$

Lemma 53.3.5. Let $k$ be a field. Let $X$ be a proper curve over $k$ with $H^0(X, \mathcal{O}_ X) = k$. If $X$ has a $\mathfrak g^ r_ d$, then $r \leq d$. If equality holds, then $H^1(X, \mathcal{O}_ X) = 0$, i.e., the genus of $X$ (Definition 53.8.1) is $0$.

**Proof.**
Let $(\mathcal{L}, V)$ be a $\mathfrak g^ r_ d$. Since this will only increase $r$, we may assume $V = H^0(X, \mathcal{L})$. Choose a nonzero element $s \in V$. 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. By Varieties, Lemma 33.44.9 we have $\deg (D) = \deg (\mathcal{L}) = d$. Since $D$ is an Artinian scheme we have $\mathcal{L}|_ D \cong \mathcal{O}_ D$^{1}. Thus

On the other hand, by assumption $\dim _ k H^0(X, \mathcal{O}_ X) = 1$ and $\dim H^0(X, \mathcal{L}) = r + 1$. We conclude that $r + 1 \leq 1 + d$, i.e., $r \leq d$ as in the lemma.

Assume equality holds. Then $H^0(X, \mathcal{L}) \to H^0(X, \mathcal{L}|_ D)$ is surjective. If we knew that $H^1(X, \mathcal{L})$ was zero, then we would conclude that $H^1(X, \mathcal{O}_ X)$ is zero by the long exact cohomology sequence and the proof would be complete. Our strategy will be to replace $\mathcal{L}$ by a large power which has vanishing. As $\mathcal{L}|_ D$ is the trivial invertible module (see above), we can find a section $t$ of $\mathcal{L}$ whose restriction of $D$ generates $\mathcal{L}|_ D$. Consider the multiplication map

and consider the short exact sequence

Since $H^0(\mathcal{L}) \to H^0(\mathcal{L}|_ D)$ is surjective and since $t$ maps to a trivialization of $\mathcal{L}|_ D$ we see that $\mu (H^0(X, \mathcal{L}) \otimes t)$ gives a subspace of $H^0(X, \mathcal{L}^{\otimes 2})$ surjecting onto the global sections of $\mathcal{L}^{\otimes 2}|_ D$. Thus we see that

Ok, so $\mathcal{L}^{\otimes 2}$ has the same property as $\mathcal{L}$, i.e., that the dimension of the space of global sections is equal to the degree plus one. Since $\mathcal{L}$ is ample (Varieties, Lemma 33.44.14) there exists some $n_0$ such that $\mathcal{L}^{\otimes n}$ has vanishing $H^1$ for all $n \geq n_0$ (Cohomology of Schemes, Lemma 30.16.1). Thus applying the argument above to $\mathcal{L}^{\otimes n}$ with $n = 2^ m$ for some sufficiently large $m$ we conclude the lemma is true. $\square$

Remark 53.3.6 (Classical definition). Let $X$ be a smooth projective curve over an algebraically closed field $k$. We say two effective Cartier divisors $D, D' \subset X$ are *linearly equivalent* if and only if $\mathcal{O}_ X(D) \cong \mathcal{O}_ X(D')$ as $\mathcal{O}_ X$-modules. Since $\mathop{\mathrm{Pic}}\nolimits (X) = \text{Cl}(X)$ (Divisors, Lemma 31.27.7) we see that $D$ and $D'$ are linearly equivalent if and only if the Weil divisors associated to $D$ and $D'$ define the same element of $\text{Cl}(X)$. Given an effective Cartier divisor $D \subset X$ of degree $d$ the *complete linear system* or *complete linear series* $|D|$ of $D$ is the set of effective Cartier divisors $E \subset X$ which are linearly equivalent to $D$. Another way to say it is that $|D|$ is the set of closed points of the fibre of the morphism

(Picard Schemes of Curves, Lemma 44.6.7) over the closed point corresponding to $\mathcal{O}_ X(D)$. This gives $|D|$ a natural scheme structure and it turns out that $|D| \cong \mathbf{P}^ m_ k$ with $m + 1 = h^0(\mathcal{O}_ X(D))$. In fact, more canonically we have

where $(-)^\vee $ indicates $k$-linear dual and $\mathbf{P}$ is as in Constructions, Example 27.21.2. In this language a *linear system* or a *linear series* on $X$ is a closed subvariety $L \subset |D|$ which can be cut out by linear equations. If $L$ has dimension $r$, then $L = \mathbf{P}(V^\vee )$ where $V \subset H^0(X, \mathcal{O}_ X(D))$ is a linear subspace of dimension $r + 1$. Thus the classical linear series $L \subset |D|$ corresponds to the linear series $(\mathcal{O}_ X(D), V)$ as defined above.

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