Break up f into its irreducible components:

f

= f1 ft

which likewise breaks Y into its irreducible components:

Y

= Z(f1) ∪ ∪ Z(ft)

Let P ∈ Z(f₁) ∩ Z(f₂), and assume without loss of generality that (P)≠0 (as given by the condition). Note that, using the product rule,

	= ⋅ f2 ft + ⋅ f1
(P)	= (P) ⋅ f2(P) ft(P) + (P) ⋅ f1(P)
	= 0
	(since f1(P) = 0,f2(P) = 0

But this contradicts the nonzeroness of the partial at P. Hence Z(f₁) ∩ Z(f₂) = ∅. But this contradicts 3.7. Hence Y must only have one irreducible component i.e. Y is irreducible (i.e. f is irreducible). Now that we know Y is a projective variety, we can apply what we learned yesterday ("Paddle Person"... seriously? Fuck my life. I deserve to die). Since the gradient vector (Jacobian) has a nonzero partial derivative at every point, we're dun.