Computing the degree of a projection map from a singular point on a quartic in $ \mathbb{P}^{4}_{\mathbb{C}} $ to $ \mathbb{P}^{3}_{C}. $ Definition:  If $ X,Y $ are irreducible varieties over $ \mathbb{C}, $ and $ f: X \rightarrow Y $ is regular map such that $ f(X) \subset Y $ is dense, then $$ \operatorname{deg}f = [\mathbb{C}(X):f^{*}\mathbb{C}(Y)]. $$
Let $ V $ be a quartic hypersurface in $ \mathbb{P}^{4} $ and $ \pi_{p}: V \rightarrow \mathbb{P}^{3} $ be the projection from some singular point $ p $ on $ V. $ I know that $ \pi_{p} $ is of degree 2, and I am trying to show this but I am stuck at an early hurdle.
$$ \operatorname{deg}\pi_{p} = [\mathbb{C}(V):\pi_{p}^{*}(\mathbb{C}(\mathbb{P}^{3}))]  =  [\mathbb{C}(V):\pi_{p}^{*}(\mathbb{C}(x_{0}:x_{1}:x_{2}:x_{3}))]  $$
I need to use some kind of general parameterisation for the points on $ V $ to procceed, but I don't know how to construct it without restricting myself to a specific example. Perhaps there is some known property that explains this, so that one does not need to calculate this directly.
 A: First we may assume that $V$ is irreducible in order to have a function field.
Up to a linear change in homogeneous coordinates we may assume that $p=(0:0:0:0:1)$ and that $V$ is given by 
$$
F(x_0,x_1,x_2,x_3,x_4) = A_4(x_0,x_1,x_2,x_3) + A_3(x_0,x_1,x_2,x_3)x_4 + A_2(x_0,x_1,x_2,x_3)x_4^2+ A_1(x_0,x_1,x_2,x_3)x_4^3 + A_0(x_0,x_1,x_2,x_3)x_4^4
$$
where $A_j$ is a homogheneous polynomial of degree $j$. Since $p\in V$ then $A_0 =0$ and since $V$ is singular at $p$ we also have $A_1=0$. Then 
$$
F(x_0,x_1,x_2,x_3,x_4) = A_4(x_0,x_1,x_2,x_3) + A_3(x_0,x_1,x_2,x_3)x_4 + A_2(x_0,x_1,x_2,x_3)x_4^2.
$$
Suppose that $A_2 \neq 0$.
Now we turn to the function fields. We first restrict to the affine charts $\{x_0=1\}$ so that $\mathbb{C}(\mathbb{P}^3) = \mathbb{C}(x_1,x_2,x_3)$  and $\mathbb{C}(V)$ is the fraction field of the domain $\dfrac{\mathbb{C}[x_1,x_2,x_3,x_4]}{(F(1,x_1,x_2,x_3,x_4))}$. It follows that 
$$\mathbb{C}(V) = \mathbb{C}(x_1,x_2,x_3)(\alpha)$$
where $\dfrac{F(1,x_1,x_2,x_3,x_4)}{A_2(1,x_1,x_2,x_3)}$ is a minimal polynomial for $\alpha$ over $\mathbb{C}(x_1,x_2,x_3)$ -- indeed it is irreducible. Hence 
$$
\deg \pi_p|_V = [\mathbb{C}(V):\mathbb{C}(\mathbb{P}^3)] = \deg_{x_4}\dfrac{F(1,x_1,x_2,x_3,x_4)}{A_2(1,x_1,x_2,x_3)} = 2
$$
To finnish note that if $A_2 =0$ and $A_3 \neq 0$ then $\pi_p$ has degree $1$, it is a birational map whose inverse is $(x_0:x_1:x_2:x_3) \mapsto \left(x_0:x_1:x_2:x_3:\dfrac{-A_4}{A_3}\right)$. If $A_3=A_4 =0$ then $V$ is a cone over $\{A_4=0\}\subset \mathbb{P}^3$ with vertex $p$. In particular $\pi_p$ is not dominant hence the morphism $\pi_p^\ast \colon \mathbb{C}(\mathbb{P}^3) \rightarrow \mathbb{C}(V)$ is not defined.
