There are at least 4 global sections on a complete intersection curve This is Vakil 19.8 C, self-study.
We take a regular complete intersection of a quadric $Q$ and a cubic $K$ in $\mathbb P^3_k$, producing a genus $4$ curve $C$. The problem asks us to show that there are at least 4 sections of $\mathcal O_C(1)$, i.e.
$$h^0(C, \mathcal O_C(1)) \geq 4$$
and says that this translates to showing that $C$ does not lie in a hyperplane. I do not see why either of these statements is true.
First, showing $C$ is not in a hyperplane is trivial it seems. If it were in a degree $d$ hypersurface, the hypersurface would have to look like $f_1Q + f_2K$, where $f_i$ are homogeneous polynomials of the correct degree. This is impossible if $d = 1$.
Second, if I ignore the translated problem and focus on the original problem, the only thing I can think to use is Riemann-Roch and Serre Duality in combination:
$$h^0(\mathcal O_C(1)) = \deg \mathcal O_C(1) - g + 1 + h^1(\mathcal O_C(1)) = 1 - 4 + 1 + h^0(\omega_C \otimes \mathcal O_C(-1))$$
but I do not see how to show that last term is $\geq 6$. We know $h^0(\omega_C) = 4$, for what it's worth.
 A: Since $C$ is a complete intersection, $\deg C = 2\cdot 3 = 6$, which means that actually $\deg \mathcal O_C(1) = 6$ (i.e. $C$ intersects a general hyperplane in $6$ points) giving us $h^0(\mathcal O_C(1)) = 6 - 4 + 1 + h^0(\omega_C \otimes \mathcal O_C(-1))$.
To determine this last term (which needs to be $\geq 1$), note that $\deg \omega_C = 6$ so the tensor product is a line bundle of degree $0$, hence has a section if and only if it is isomorphic to $\mathcal O_C$ which is itself the case if and only if $\mathcal O_C(1) \cong \omega_C$. This follows from iterated adjunction:
$$
\omega_Q \cong \omega_{\mathbb P^3}\otimes \mathcal O_{\mathbb P^3}(Q)|_Q \cong \mathcal O_{\mathbb P^3}(-4)\otimes\mathcal O_{\mathbb P^3}(2)|_Q \cong \mathcal O_{\mathbb P^3}(-2)|_Q,
$$
and then
$$
\omega_C \cong \omega_Q\otimes (\mathcal O_{\mathbb P^3}(K)|_Q)|_C \cong \mathcal O_{\mathbb P^3}(-2)|_Q \otimes (\mathcal O_{\mathbb P^3}(3)|_Q)|_C \cong \mathcal O_{\mathbb P^3}(1)|_C = \mathcal O_C(1),
$$
where I've maybe gone a bit overboard with the restriction notation, but hopefully the idea is clear: everything happening is with line bundles that have been restricted from projective space.
