# Very simple characterisation of $(-1)$-curves

Let $$X$$ be a complex surface (i.e., complex dimension 2) and $$C$$ a $$(-1)$$-curve in $$X$$, i.e., a reduced, compact, connected curve $$C$$ with self-intersection $$-1$$. I'm trying to understand the proof of the equivalent characterization of $$(-1)$$-curves, namely, $$C^2 < 0 \ \ \text{and} \ \ (C, K_X) < 0.$$ Indeed, if $$C$$ is a $$(-1)$$-curve, then $$C^2 =-1$$ by definition and by adjunction we have $$K_C = K_X \otimes \mathcal{O}_X(C) \vert_C.$$ Then $$(C, K_C) = (C, K_X) + (C, \mathcal{O}_X(C) \vert_C),$$ since the intersection product is bilinear with respect to the tensor product of line bundles. How do we deduce from this that $$(C,K_X) <0?$$

The adjunction formula allows us to write $$\text{deg}(K_C) = (K_X + C)C = (K_X \cdot C) + C^2.$$ Now we use that $$C$$ is a $$(-1)$$-curve:
• It is isomorphic to $$\mathbb{P}^1$$, so that $$\text{deg}(K_C) = -2$$
• We have $$C^2 = -1$$
Hence we get $$-2 = (K_X \cdot C) -1$$ by the formula above, i.e. $$(K_X \cdot C) = -1$$.