Let $X$ be a compact complex Kähler manifold and $L$ an holomorphic line bundle over $X$. Let suppose $c_{1}(L) = 0$ in $H^{1, 1}(X, \mathbb{C})$.
I would like to show that if $L$ is not the trivial bundle, then $L$ doesn't admit any non trivial holomorphic section.
Idea : -I can't use the usual vanishing theorems (Serre or Koddaira).
-If $L$ would admit such a section $s$, I'd like to show it vanishes nowhere. However, I don't have hypothesis on $s^{-1}(\{0\})$.
I wish you a good day.
In Székelyhidi's book (in the paragraph including equation 1.17) it is shown that a line bundle $L$ such that $c_1(L)<0$ does not admit any nontrivial holomorphic section. The same procedure as in the proof of that result can be used to solve Exercise 1.42, I will try to explain this below. I would also like to propose an alternative reasoning, tell me what you think.
As $c_1(L)=0$, $L$ admits a Chern-flat Hermitian metric $h$. Then, for the Chern connection we have $\nabla_a\nabla_{\bar{b}}=\nabla_{\bar{b}}\nabla_a$. If $s\in H^0(L)$ then it follows that $\nabla_{\bar{b}}\nabla_as=0$. Integrating by parts as in Székelyhidi's book we then find $\left\langle\nabla^{1,0} s,\nabla^{1,0} s\right\rangle_{L^2}=0$. Hence any holomorphic section of $L$ is covariantly constant. In particular, this implies that $\lvert s\rvert^2_h$ is constant. So, if $s$ vanishes somewhere it must vanish everywhere, proving the claim.
Alternatively, consider this reasoning: the curvature form of $h$ vanishes, and this curvature is locally expressed as $-i\,\partial\bar{\partial}\log h(s_0,s_0)$ for any local holomorphic frame of $L$. Now, fix $s\in H^0(L)$ and assume that $s$ does not vanish identically. Then, the set $Z=\{s=0\}$ is a divisor in $M$. Outside of $Z$, the curvature form of $h$ then is given by $-i\,\partial\bar{\partial}\log\lvert s\rvert^2_h$, and we are assuming that this is identically zero. So, consider $f:=\log\lvert s\rvert^2_h$. This is a well-defined function on $M_0=M\setminus Z$, harmonic in $M_0$, and $f$ goes to $-\infty$ at $Z$. The only way not to contradict the maximum principle is for $f$ to be constant, and this allows us to conclude as before.