Lemma 53.9.3. Let $Z \subset \mathbf{P}^2_ k$ be as in Lemma 53.9.1 and let $I(Z) = (F)$ for some $F \in k[T_0, T_1, T_2]$. Then $H^0(Z, \mathcal{O}_ Z) = k$ and the genus of $Z$ is $(d - 1)(d - 2)/2$ where $d = \deg (F)$.

Proof. Let $S = k[T_0, T_1, T_2]$. There is an exact sequence of graded modules

$0 \to S(-d) \xrightarrow {F} S \to S/(F) \to 0$

Denote $i : Z \to \mathbf{P}^2_ k$ the given closed immersion. Applying the exact functor $\widetilde{\ }$ (Constructions, Lemma 27.8.4) we obtain

$0 \to \mathcal{O}_{\mathbf{P}^2_ k}(-d) \to \mathcal{O}_{\mathbf{P}^2_ k} \to i_*\mathcal{O}_ Z \to 0$

because $F$ generates the ideal of $Z$. Note that the cohomology groups of $\mathcal{O}_{\mathbf{P}^2_ k}(-d)$ and $\mathcal{O}_{\mathbf{P}^2_ k}$ are given in Cohomology of Schemes, Lemma 30.8.1. On the other hand, we have $H^ q(Z, \mathcal{O}_ Z) = H^ q(\mathbf{P}^2_ k, i_*\mathcal{O}_ Z)$ by Cohomology of Schemes, Lemma 30.2.4. Applying the long exact cohomology sequence we first obtain that

$k = H^0(\mathbf{P}^2_ k, \mathcal{O}_{\mathbf{P}^2_ k}) \longrightarrow H^0(Z, \mathcal{O}_ Z)$

is an isomorphism and next that the boundary map

$H^1(Z, \mathcal{O}_ Z) \longrightarrow H^2(\mathbf{P}^2_ k, \mathcal{O}_{\mathbf{P}^2_ k}(-d)) \cong k[T_0, T_1, T_2]_{d - 3}$

is an isomorphism. Since it is easy to see that the dimension of this is $(d - 1)(d - 2)/2$ the proof is finished. $\square$

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