Complete intersection in $\mathbb{P}^{3}$ I was doing an exercise in Hartshorne's book:
If $X$ is a complete intersection of surfaces of degrees $a,b$ in $\mathbb{P}^{3}$, then $X$ does not lie on any surface of degree < $ \min(a,b)$.
We have the exact sequence :$0 \to \mathscr{I}_{X}(m) \to \mathcal{O}_{\mathbb{P}^{3}}(m) \to \mathcal{O}_{X}(m) \to 0$, where $m < \min(a,b)$. And I know how to find $\dim H^{0}(\mathcal{O}_{\mathbb{P}^{3}}(m))$, also $H^{0}(\mathcal{O}_{\mathbb{P}^{3}}(m)) \to H^{0}(\mathcal{O}_{X}(m))$ is surjective. But how to compute $\dim H^{0}(\mathcal{O}_{X}(m))$?
Thank you!
 A: As you say, the strategy is to consider the short exact sequence
$$
  0 \longrightarrow \mathscr{I}_X(n) \longrightarrow \mathcal{O}_{\mathbf{P}^3}(n) \longrightarrow \mathcal{O}_X(n) \longrightarrow 0
$$
for every $n \in \mathbf{Z}$ and its associated long exact sequence
$$
  0 \longrightarrow H^0(\mathscr{I}_X(n)) \longrightarrow H^0(\mathcal{O}_{\mathbf{P}^3}(n)) \longrightarrow H^0(\mathcal{O}_X(n)) \longrightarrow \cdots.
$$
We want to show that if $n < \min(a,b)$, then
$$
    H^0(\mathscr{I}_X(n)) = 0,
$$
for this would show that no degree $n$ homogeneous polynomial vanishes on all of $X$, hence $X$ is not contained in a hypersurface of degree $n$.
Consider the short exact sequence of graded modules:
$$
  0 \longrightarrow S(-a-b) \longrightarrow S(-a) \oplus S(-b) \longrightarrow I_X \longrightarrow 0
$$
which is the (graded) Koszul resolution of $I_X$, the vanishing ideal of $X$. Note the Koszul resolution is exact because $X$ is a complete intersection, hence the generators of $I_X$ form a regular sequence; see Prop. III.7.10A. Taking associated sheaves gives the short exact sequence of sheaves
$$
  0 \longrightarrow \mathcal{O}_{\mathbf{P}^3}(-a-b) \longrightarrow \mathcal{O}_{\mathbf{P}^3}(-a) \oplus \mathcal{O}_{\mathbf{P}^3}(-b) \longrightarrow \mathscr{I}_X \longrightarrow 0,
$$
and after twisting by $n$ and taking the associated long exact sequence on cohomology, we obtain the exact sequence
$$
  0 = H^0(\mathcal{O}_{\mathbf{P}^3}(n-a)) \oplus H^0(\mathcal{O}_{\mathbf{P}^3}(n-b)) \longrightarrow H^0(\mathscr{I}_X(n)) \longrightarrow H^1(\mathcal{O}_{\mathbf{P}^3}(n-a-b)) = 0
$$
as long as $n < \min(a,b)$ by using the cohomology of $\mathbf{P}^3$ (Thm. III.5.1). Thus, $H^0(\mathscr{I}_X(n)) = 0$ if $n < \min(a,b)$, and so $X$ is not contained in any surface of degree $n < \min(a,b)$. $\blacksquare$
Remark. It's a bit unclear to me how computing $\dim H^0(\mathcal{O}_X(n))$ is useful, but maybe I am missing something. A result coming out of this proof is that the restriction map $H^0(\mathcal{O}_{\mathbf{P}^3}(n)) \to H^0(\mathcal{O}_X(n))$ is injective in the range $n < \min(a,b)$.
