← I.5.8 I.5.10bc →



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 ∂f-
 ∂x(P)0 (as given by the condition). Note that, using the product rule,

∂x = ∂f1--
 ∂x f2 ⋅⋅⋅ft + ∂f2-⋅-⋅⋅ft-
    ∂x f1
= ⇒∂f--
∂x(P) = ∂f1--
 ∂x(P) f2(P) ⋅⋅ ⋅ft(P) + ∂f2-⋅⋅⋅-ft-
   ∂x(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.

← I.5.8 I.5.10bc →