Problem with the definition of the degree of a morphism between curves I need your help because there is something that I don't understand in the definition of the degree of a morphism of curves. Here it is:
"Let $\phi: C_1 \longrightarrow C_2$ be a morphism between two (smooth) projective curves. We define the degree of $\phi$ to be 0 if it is constant and $deg\phi=[k(C_1):\phi^*k(C_2])$ otherwise."
I don't understand why the curves can be singular. The point is, the degree is only defined by the fraction fields of the curves, but why does the smoothness have no "impact" on them?
Thanks.
 A: Here I assume that the ground field is algebraically closed for simplicity but this is not essential. When a morphism of curves $\phi : C_1 \to C_2$ is finite and flat then you can show that ``the degree of each fiber'' equals the degree defined as $\deg{\phi} = [k(C_1) : k(C_2)]$. Explicitly, this means for each $Q \in C_2$,
$$ \sum_{P \in f^{-1}(Q)} e_P = \deg{\phi} $$
where $e_P$ is the ramification degree at $P$. You can think of $[k(C_1) : k(C_2)]$  as the degree of the generic fiber.
Now, any nonconstant morphism of projective curves is automatically finite and if $C_2$ is smooth it is also automatically flat. However, when $C_2$ is not smooth it does not need to be flat and accordingly the above formula may not hold.
For example, let $X$ be the nodal cubic $y^2 = x^2(x - 1)$ in $\mathbb{P}^2$. Then the normalization is a map $\nu : \mathbb{P}^1 \to X$ defined by $t \mapsto (t^2 + 1, t(t^2 + 1))$. You can show this morphism is an isomorphism away from the node at $(0,0)$ so $\deg{\nu} = 1$. However, $\pm i$ both map to $(0,0)$ so the degree over the node is $2$.
