The Stacks project

Proof. More precisely, set $X_{k'} = X \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(k')$. Let $\mathcal{E}_{k'} = p^*\mathcal{E}$ where $p : X_{k'} \to X$ is the projection. By Cohomology of Schemes, Lemma 30.5.2 we have $H^ i(X_{k'}, \mathcal{E}_{k'}) = H^ i(X, \mathcal{E}) \otimes _ k k'$ and $H^ i(X_{k'}, \mathcal{O}_{X_{k'}}) = H^ i(X, \mathcal{O}_ X) \otimes _ k k'$. Hence we see that the Euler characteristics are unchanged, hence the degree is unchanged. $\square$

Proof. Follows immediately from additivity of Euler characteristics (Lemma 33.33.2) and additivity of ranks. $\square$

Proof. The morphism $f$ is proper (Morphisms, Lemma 29.41.7) and has fibres of dimension $\leq 0$. Hence $f$ is finite (Cohomology of Schemes, Lemma 30.21.2). Thus

Since $f$ induces an isomorphism on local rings at generic points of all irreducible components of dimension $1$ we see that the kernel and cokernel

have supports of dimension $\leq 0$. Note that tensoring this with $\mathcal{E}$ is still an exact sequence as $\mathcal{E}$ is locally free. We obtain

which proves what we want. The first equality as $f$ is finite, see Cohomology of Schemes, Lemma 30.2.4. The second equality by projection formula, see Cohomology, Lemma 20.54.2. The third by additivity of Euler characteristics, see Lemma 33.33.2. The fourth by Lemma 33.33.3. $\square$

Proof. Let $\mathcal{P}$ be the property of coherent sheaves $\mathcal{F}$ on $X$ expressing that the formula of the lemma holds. We claim that the assumptions (1) and (2) of Cohomology of Schemes, Lemma 30.12.6 hold for $\mathcal{P}$. Namely, (1) holds because the Euler characteristic and the rank $r$ are additive in short exact sequences of coherent sheaves. And (2) holds too: If $Z = X$ then we may take $\mathcal{G} = \mathcal{O}_ X$ and $\mathcal{P}(\mathcal{O}_ X)$ is true by the definition of degree. If $i : Z \to X$ is the inclusion of a closed point we may take $\mathcal{G} = i_*\mathcal{O}_ Z$ and $\mathcal{P}$ holds by Lemma 33.33.3 and the fact that $r = 0$ in this case. $\square$

Proof. Observe that the statement makes sense because $C_ i \to \mathop{\mathrm{Spec}}(k)$ is proper of dimension $1$ (Morphisms, Lemmas 29.41.6 and 29.41.4). Consider the open subscheme $U_ i = X \setminus (\bigcup _{j \not= i} C_ j)$ and let $X_ i \subset X$ be the scheme theoretic closure of $U_ i$. Note that $X_ i \cap U_ i = U_ i$ (scheme theoretically) and that $X_ i \cap U_ j = \emptyset $ (set theoretically) for $i \not= j$; this follows from the description of scheme theoretic closure in Morphisms, Lemma 29.7.7. Thus we may apply Lemma 33.44.4 to the morphism $X' = \bigcup X_ i \to X$. Since it is clear that $C_ i \subset X_ i$ (scheme theoretically) and that the multiplicity of $C_ i$ in $X_ i$ is equal to the multiplicity of $C_ i$ in $X$, we see that we reduce to the case discussed in the following paragraph.

Assume $X$ is irreducible with generic point $\xi $. Let $C = X_{red}$ have multiplicity $m$. We have to show that $\deg (\mathcal{E}) = m \deg (\mathcal{E}|_ C)$. Let $\mathcal{I} \subset \mathcal{O}_ X$ be the ideal defining the closed subscheme $C$. Let $e \geq 0$ be minimal such that $\mathcal{I}^{e + 1} = 0$ (Cohomology of Schemes, Lemma 30.10.2). We argue by induction on $e$. If $e = 0$, then $X = C$ and the result is immediate. Otherwise we set $\mathcal{F} = \mathcal{I}^ e$ viewed as a coherent $\mathcal{O}_ C$-module (Cohomology of Schemes, Lemma 30.9.8). Let $X' \subset X$ be the closed subscheme cut out by the coherent ideal $\mathcal{I}^ e$ and let $m'$ be the multiplicity of $C$ in $X'$. Taking stalks at $\xi $ of the short exact sequence

we find (use Algebra, Lemmas 10.52.3, 10.52.6, and 10.52.5) that

where $r$ is the rank of $\mathcal{F}$ as a coherent sheaf on $C$. Tensoring with $\mathcal{E}$ we obtain a short exact sequence

By induction we have $\chi (\mathcal{E} \otimes \mathcal{O}_{X'}) = m' \deg (\mathcal{E}|_ C)$. By Lemma 33.44.5 we have $\chi (\mathcal{E}|_ C \otimes \mathcal{F}) = r \deg (\mathcal{E}|_ C) + n \chi (\mathcal{F})$. Putting everything together we obtain the result. $\square$

Proof. By Lemma 33.44.6 and elementary arithmetic, we reduce to the case of a proper curve. This case follows from Lemma 33.44.5. $\square$

Proof. By Lemma 33.44.6 and elementary arithmetic, we reduce to the case of a proper curve. Then there exists a modification $f : X' \to X$ such that $f^*\mathcal{E}$ has a filtration whose successive quotients are invertible modules, see Divisors, Lemma 31.36.1. By Lemma 33.44.4 we may work on $X'$. Thus we may assume we have a filtration

by locally free $\mathcal{O}_ X$-modules with $\mathcal{L}_ i = \mathcal{E}_ i/\mathcal{E}_{i - 1}$ is invertible. By Modules, Lemma 17.26.1 and induction we find $\det (\mathcal{E}) = \mathcal{L}_1 \otimes \ldots \otimes \mathcal{L}_ n$. Thus the equality follows from Lemma 33.44.7 and additivity (Lemma 33.44.3). $\square$

Proof. Since $D$ is nowhere dense in $X$ (Divisors, Lemma 31.13.4) we see that $\dim (D) \leq 0$. Hence $D$ is finite over $k$ by Lemma 33.20.2. Since $k$ is a field, the morphism $D \to \mathop{\mathrm{Spec}}(k)$ is finite locally free and hence has a degree (Morphisms, Definition 29.48.1), which is clearly equal to $\dim _ k \Gamma (D, \mathcal{O}_ D)$ as stated in the lemma. By Divisors, Definition 31.14.1 there is a short exact sequence

where $i : D \to X$ is the closed immersion. Tensoring with $\mathcal{E}$ we obtain a short exact sequence

The equation of the lemma follows from additivity of the Euler characteristic (Lemma 33.33.2) and Lemma 33.33.3. $\square$

Proof. See Lemma 33.9.3 for the assertions about $\kappa $. For every quasi-coherent $\mathcal{O}_ X$-module, the $k$-vector spaces $H^ i(X, \mathcal{F})$ are $\kappa $-vector spaces. The divisibilities easily follow from this statement and the definitions. $\square$

Proof. The degree of $X$ over $Y$ is defined in Morphisms, Definition 29.51.8. Thus $f_*\mathcal{O}_ X$ is a coherent $\mathcal{O}_ Y$-module of rank $\deg (X/Y)$, i.e., $\deg (X/Y) = \dim _{\kappa (\xi )} (f_*\mathcal{O}_ X)_\xi $ where $\xi $ is the generic point of $Y$. Thus we obtain

as desired. The first equality as $f$ is finite, see Cohomology of Schemes, Lemma 30.2.4. The second equality by projection formula, see Cohomology, Lemma 20.54.2. The third equality by Lemma 33.44.5. $\square$

Proof. Let $s$ be a nonzero section of $\mathcal{L}$. Since $X$ is a curve, we see that $s$ is a regular section. Hence there is an effective Cartier divisor $D \subset X$ and an isomorphism $\mathcal{L} \to \mathcal{O}_ X(D)$ mapping $s$ the canonical section $1$ of $\mathcal{O}_ X(D)$, see Divisors, Lemma 31.14.10. Then $\deg (\mathcal{L}) = \deg (D)$ by Lemma 33.44.9. As $\deg (D) \geq 0$ and $= 0$ if and only if $D = \emptyset $, this proves (1) and (2). In case (3) we see that $\deg (\mathcal{L}) = 0$ and $D = \emptyset $. Similarly for (4). To see (5) apply (1) and (4) to the invertible sheaf

which has degree $\deg (\mathcal{L}) - \deg (\mathcal{N})$ by Lemma 33.44.7. $\square$

Proof. Let $s \in H^0(X, \mathcal{L})$ be nonzero. Since $X$ is reduced there exists an irreducible component $C$ of $X$ with $s|_ C \not= 0$. But if $s|_ C$ is nonzero, then $s$ is nonwhere vanishing on $C$ by Lemma 33.44.12. This in turn implies $s$ is nowhere vanishing on every irreducible component of $X$ meeting $C$. Since $X$ is connected, we conclude that $s$ vanishes nowhere and the lemma follows. $\square$

Proof. If $\mathcal{L}$ is ample, then there exists an $n > 0$ and a section $s \in H^0(X, \mathcal{L}^{\otimes n})$ with $X_ s$ affine. Since $X$ isn't affine (otherwise by Morphisms, Lemma 29.44.11 $X$ would be finite), we see that $s$ vanishes at some point. Hence $\deg (\mathcal{L}^{\otimes n}) > 0$ by Lemma 33.44.12. By Lemma 33.44.7 we conclude that $\deg (\mathcal{L}) = 1/n\deg (\mathcal{L}^{\otimes n}) > 0$.

Assume $\deg (\mathcal{L}) > 0$. Then

grows linearly with $n$. Hence for any finite collection of closed points $x_1, \ldots , x_ t$ of $X$, we can find an $n$ such that $\dim _ k H^0(X, \mathcal{L}^{\otimes n}) > \sum \dim _ k \kappa (x_ i)$. (Recall that by Hilbert Nullstellensatz, the extension fields $\kappa (x_ i)/k$ are finite, see for example Morphisms, Lemma 29.20.3). Hence we can find a nonzero $s \in H^0(X, \mathcal{L}^{\otimes n})$ vanishing in $x_1, \ldots , x_ t$. In particular, if we choose $x_1, \ldots , x_ t$ such that $X \setminus \{ x_1, \ldots , x_ t\} $ is affine, then $X_ s$ is affine too (for example by Properties, Lemma 28.26.4 although if we choose our finite set such that $\mathcal{L}|_{X \setminus \{ x_1, \ldots , x_ t\} }$ is trivial, then it is immediate). The conclusion is that we can find an $n > 0$ and a nonzero section $s \in H^0(X, \mathcal{L}^{\otimes n})$ such that $X_ s$ is affine.

We will show that for every quasi-coherent sheaf of ideals $\mathcal{I}$ there exists an $m > 0$ such that $H^1(X, \mathcal{I} \otimes \mathcal{L}^{\otimes m})$ is zero. This will finish the proof by Cohomology of Schemes, Lemma 30.17.1. To see this we consider the maps

Since $\mathcal{I}$ is torsion free, these maps are injective and isomorphisms over $X_ s$, hence the cokernels have vanishing $H^1$ (by Cohomology of Schemes, Lemma 30.9.10 for example). We conclude that the maps of vector spaces

are surjective. On the other hand, the dimension of $H^1(X, \mathcal{I})$ is finite, and every element maps to zero eventually by Cohomology of Schemes, Lemma 30.17.4. Thus for some $e > 0$ we see that $H^1(X, \mathcal{I} \otimes \mathcal{L}^{\otimes en})$ is zero. This finishes the proof. $\square$

Proof. Let $x_1, \ldots , x_ r \in X$ be the isolated closed points. Think of $x_ i = \mathop{\mathrm{Spec}}(\kappa (x_ i))$ as a scheme. Consider the morphism of schemes

This is a finite surjective morphism of schemes proper over $k$ (details omitted). Thus $\mathcal{L}$ is ample if and only if $f^*\mathcal{L}$ is ample (Cohomology of Schemes, Lemma 30.17.2). Thus we conclude by Lemma 33.44.14. $\square$

Proof. Choose a closed immersion $i : X \to \mathbf{P}^ n_ k$ (Lemma 33.43.4). Setting $\mathcal{L} = i^*\mathcal{O}_{\mathbf{P}^ n}(d)$ for $d \gg 0$ we see that there exists an invertible sheaf $\mathcal{L}$ with $H^0(X, \mathcal{L}) \not= 0$ and $H^1(X, \mathcal{L}) = 0$ (see Cohomology of Schemes, Lemma 30.17.1 for vanishing and the references therein for nonvanishing). We will finish the proof of (1) by descending induction on $t = \dim _ k H^0(X, \mathcal{L})$. The base case $t = 1$ is trivial. Assume $t > 1$.

Let $U \subset X$ be the nonempty open subset of nonsingular points studied in Lemma 33.25.8. Let $s \in H^0(X, \mathcal{L})$ be nonzero. There exists a closed point $x \in U$ such that $s$ does not vanish in $x$. Let $\mathcal{I}$ be the ideal sheaf of $i : x \to X$ as in Lemma 33.43.8. Look at the short exact sequence

Observe that $H^0(X, i_*i^*\mathcal{L}) = H^0(x, i^*\mathcal{L})$ has dimension $1$ as $x$ is a $k$-rational point ($k$ is algebraically closed). Since $s$ does not vanish at $x$ we conclude that

is surjective. Hence $\dim _ k H^0(X, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}) = t - 1$. Finally, the long exact sequence of cohomology also shows that $H^1(X, \mathcal{I} \otimes _{\mathcal{O}_ X} \mathcal{L}) = 0$ thereby finishing the proof of the induction step.

To get an invertible sheaf as in (2) take an invertible sheaf $\mathcal{L}$ as in (1) and do the argument in the previous paragraph one more time. $\square$

Proof. Let $\mathcal{N}$ be the invertible module we found in Lemma 33.44.16 part (2). The degree of $\mathcal{N}$ is $\chi (X, \mathcal{N}) - \chi (X, \mathcal{O}_ X) = 0 - (1 - g) = g - 1$. Hence the degree of $\mathcal{L} \otimes \mathcal{N}^{\otimes - 1}$ is $\deg (\mathcal{L}) - (g - 1) \geq g$. Hence $\chi (X, \mathcal{L} \otimes \mathcal{N}^{\otimes -1}) \geq g + 1 - g = 1$. Thus there is a nonzero global section $s$ whose zero scheme is an effective Cartier divisor $D$ of degree $\deg (\mathcal{L}) - (g - 1)$. This gives a short exact sequence

where $i : D \to X$ is the inclusion morphism. We conclude that $H^0(X, \mathcal{L})$ maps isomorphically to $H^0(D, \mathcal{L}|_ D)$ which has dimension $\deg (\mathcal{L}) - (g - 1)$. The result follows from the definition of degree. $\square$