Let $k$ be a field with algebraic closure $\overline{k}$. Let $X$ be a proper scheme of dimension $\leq 1$ over $k$. We define $g_{geom}(X/k)$ to be the sum of the geometric genera of the irreducible components of $X_{\overline{k}}$ which have dimension $1$.
Lemma 53.18.1. Let $k$ be a field. Let $X$ be a proper scheme of dimension $\leq 1$ over $k$. Then where the sum is over irreducible components $C \subset X$ of dimension $1$.
\[ g_{geom}(X/k) = \sum \nolimits _{C \subset X} g_{geom}(C/k) \]
Proof. This is immediate from the definition and the fact that an irreducible component $\overline{Z}$ of $X_{\overline{k}}$ maps onto an irreducible component $Z$ of $X$ (Varieties, Lemma 33.8.10) of the same dimension (Morphisms, Lemma 29.28.3 applied to the generic point of $\overline{Z}$). $\square$
Lemma 53.18.2. Let $k$ be a field. Let $X$ be a proper scheme of dimension $\leq 1$ over $k$. Then
We have $g_{geom}(X/k) = g_{geom}(X_{red}/k)$.
If $X' \to X$ is a birational proper morphism, then $g_{geom}(X'/k) = g_{geom}(X/k)$.
If $X^\nu \to X$ is the normalization morphism, then $g_{geom}(X^\nu /k) = g_{geom}(X/k)$.
Proof. Part (1) is immediate from Lemma 53.18.1. If $X' \to X$ is proper birational, then it is finite and an isomorphism over a dense open (see Varieties, Lemmas 33.17.2 and 33.17.3). Hence $X'_{\overline{k}} \to X_{\overline{k}}$ is an isomorphism over a dense open. Thus the irreducible components of $X'_{\overline{k}}$ and $X_{\overline{k}}$ are in bijective correspondence and the corresponding components have isomorphic function fields. In particular these components have isomorphic nonsingular projective models and hence have the same geometric genera. This proves (2). Part (3) follows from (1) and (2) and the fact that $X^\nu \to X_{red}$ is birational (Morphisms, Lemma 29.54.7). $\square$
Lemma 53.18.3. Let $k$ be a field. Let $X$ be a proper scheme of dimension $\leq 1$ over $k$. Let $f : Y \to X$ be a finite morphism such that there exists a dense open $U \subset X$ over which $f$ is a closed immersion. Then
\[ \dim _ k H^1(X, \mathcal{O}_ X) \geq \dim _ k H^1(Y, \mathcal{O}_ Y) \]
Proof. Consider the exact sequence
of coherent sheaves on $X$. By assumption $\mathcal{F}$ is supported in finitely many closed points and hence has vanishing higher cohomology (Varieties, Lemma 33.33.3). On the other hand, we have $H^2(X, \mathcal{G}) = 0$ by Cohomology, Proposition 20.20.7. It follows formally that the induced map $H^1(X, \mathcal{O}_ X) \to H^1(X, f_*\mathcal{O}_ Y)$ is surjective. Since $H^1(X, f_*\mathcal{O}_ Y) = H^1(Y, \mathcal{O}_ Y)$ (Cohomology of Schemes, Lemma 30.2.4) we conclude the lemma holds. $\square$
Lemma 53.18.4. Let $k$ be a field. Let $X$ be a proper scheme of dimension $\leq 1$ over $k$. If $X' \to X$ is a birational proper morphism, then If $X$ is reduced, $H^0(X, \mathcal{O}_ X) \to H^0(X', \mathcal{O}_{X'})$ is surjective, and equality holds, then $X' = X$.
\[ \dim _ k H^1(X, \mathcal{O}_ X) \geq \dim _ k H^1(X', \mathcal{O}_{X'}) \]
Proof. If $f : X' \to X$ is proper birational, then it is finite and an isomorphism over a dense open (see Varieties, Lemmas 33.17.2 and 33.17.3). Thus the inequality by Lemma 53.18.3. Assume $X$ is reduced. Then $\mathcal{O}_ X \to f_*\mathcal{O}_{X'}$ is injective and we obtain a short exact sequence
Under the assumptions given in the second statement, we conclude from the long exact cohomology sequence that $H^0(X, \mathcal{F}) = 0$. Then $\mathcal{F} = 0$ because $\mathcal{F}$ is generated by global sections (Varieties, Lemma 33.33.3). and $\mathcal{O}_ X = f_*\mathcal{O}_{X'}$. Since $f$ is affine this implies $X = X'$. $\square$
Lemma 53.18.5. Let $k$ be a field. Let $C$ be a proper curve over $k$. Set $\kappa = H^0(C, \mathcal{O}_ C)$. Then
\[ [\kappa : k]_ s \dim _\kappa H^1(C, \mathcal{O}_ C) \geq g_{geom}(C/k) \]
Proof. Varieties, Lemma 33.26.2 implies $\kappa $ is a field and a finite extension of $k$. By Fields, Lemma 9.14.8 we have $[\kappa : k]_ s = |\mathop{\mathrm{Mor}}\nolimits _ k(\kappa , \overline{k})|$ and hence $\mathop{\mathrm{Spec}}(\kappa \otimes _ k \overline{k})$ has $[\kappa : k]_ s$ points each with residue field $\overline{k}$. Thus
(set theoretic union). Here $C_ t = C \times _{\mathop{\mathrm{Spec}}(\kappa ), t} \mathop{\mathrm{Spec}}(\overline{k})$ where we view $t$ as a $k$-algebra map $t : \kappa \to \overline{k}$. The conclusion is that $g_{geom}(C/k) = \sum _ t g_{geom}(C_ t/\overline{k})$ and the sum is over an index set of size $[\kappa : k]_ s$. We have
by cohomology and base change (Cohomology of Schemes, Lemma 30.5.2). Observe that the normalization $C_ t^\nu $ is the disjoint union of the nonsingular projective models of the irreducible components of $C_ t$ (Morphisms, Lemma 29.54.6). Hence $\dim _{\overline{k}} H^1(C_ t^\nu , \mathcal{O}_{C_ t^\nu })$ is equal to $g_{geom}(C_ t/\overline{k})$. By Lemma 53.18.3 we have
and this finishes the proof. $\square$
Lemma 53.18.6. Let $k$ be a field. Let $X$ be a proper scheme of dimension $\leq 1$ over $k$. Let $\ell $ be a prime number invertible in $k$. Then where $g_{geom}(X/k)$ is as defined above.
\[ \dim _{\mathbf{F}_\ell } \mathop{\mathrm{Pic}}\nolimits (X)[\ell ] \leq \dim _ k H^1(X, \mathcal{O}_ X) + g_{geom}(X/k) \]
Proof. The map $\mathop{\mathrm{Pic}}\nolimits (X) \to \mathop{\mathrm{Pic}}\nolimits (X_{\overline{k}})$ is injective by Varieties, Lemma 33.30.3. By Cohomology of Schemes, Lemma 30.5.2 $\dim _ k H^1(X, \mathcal{O}_ X)$ equals $\dim _{\overline{k}} H^1(X_{\overline{k}}, \mathcal{O}_{X_{\overline{k}}})$. Hence we may assume $k$ is algebraically closed.
Let $X_{red}$ be the reduction of $X$. Then the surjection $\mathcal{O}_ X \to \mathcal{O}_{X_{red}}$ induces a surjection $H^1(X, \mathcal{O}_ X) \to H^1(X, \mathcal{O}_{X_{red}})$ because cohomology of quasi-coherent sheaves vanishes in degrees $\geq 2$ by Cohomology, Proposition 20.20.7. Since $X_{red} \to X$ induces an isomorphism on irreducible components over $\overline{k}$ and an isomorphism on $\ell $-torsion in Picard groups (Picard Schemes of Curves, Lemma 44.7.2) we may replace $X$ by $X_{red}$. In this way we reduce to Proposition 53.17.3. $\square$
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