Consider the $n$-th Hirzebruch surface $\mathbb F_n:=\mathbb P_{\mathbb P^1}(\mathcal O_{\mathbb P^1}\otimes\mathcal O_{\mathbb P^1} (n))$, and let $C_0\subseteq \mathbb F_n$ a section of the ruling morphism onto $\mathbb P^1$ (this section always exists by Tsen theorem).
Theorem: Let $n>0$, then $\mathbb F_n$ contains a unique irreducible curve $C$ such that $C^2<0$. In particular $C^2=-n$.
I have two questions about this theorem:
1. I Think that the statement should be :
"...then $\mathbb F_n$ contains a unique curve $C$ up to linear equivalence such that..."
Am I right? This because if $B$ is an irreducible curve linearly equivalent to $C$ then $B^2=C^2$.
2. The proof is simple indeed one verifies that $C=C_0-nF$ where $F$ is any fiber of the ruling morphism. However I don't understand why $C_0-nF$ is irreducible. The curves $C_0$ and $F$ seem to be its irreducible components.
The curve $C$ is unique. As you say, of course if $B$ is a curve linearly equivalent to $C$ then $B^2=C^2$. However, in this case by Riemann-Roch you find $h^0(\Bbb{F}_n,\mathcal{O}( C ))=1$ and therefore the linear system $|C|$ (of curves equivalent to $C$) actually contains $C$ only. As for your second question maybe it will be clearer if you write $C_0=C+nF$ and realize that it is $C_0$ which is reducible, with $C$ and $F$ being its irreducible components.