Let $X$ be a smooth projective variety over $\mathbb{C}$, $\mathcal{L}$ be a line bundle on it. Let $s,t$ be two elements of $H^{0}(X,\mathcal{L})$, that is $s,t$ are global sections of $\mathcal{L}$. We could define the following map $$\phi:\mathcal{O}_{X}\oplus\mathcal{O}_{X}\to \mathcal{L}$$where $\phi(1,0)=s, \phi(0,1)=t$.
Now suppose $s,t$ generate $\mathcal{L}$, then we can get $\phi$ is surjective. I want to know what is the kernel of this map.
It is easy to show the kernel is also a line bundle. I guess the kernel should be $\mathcal{L}^{-1}$. But I do not know how to show it. Could you help me?
Let $\mathcal{L}'$ denite the kernel, so that there is an exact sequence $$ 0 \to \mathcal{L}' \stackrel\psi\to \mathcal{O}_X \oplus \mathcal{O}_X \stackrel\phi\to \mathcal{L} \to 0. $$ Then taking determinants gives $$ \mathcal{L}' \otimes \mathcal{L} \cong \det(\mathcal{O}_X \oplus \mathcal{O}_X) \cong \mathcal{O}_X $$ which implies $\mathcal{L}' \cong \mathcal{L}^{-1}$.
In fact, the kernel morphism can be written as $\psi = (t,-s)$.