I.5.9

8/20/2021

Break up f into its irreducible components:

f | = f_{1}
f_{t} |

which likewise breaks Y into its irreducible components:

Y | = Z(f_{1}) ∪
∪ Z(f_{t}) |

Let P ∈ Z(f_{1}) ∩ Z(f_{2}), and assume without loss of generality that
(P)≠0 (as given by the condition). Note that, using the product rule,

=
⋅ f_{2}
f_{t} +
⋅ f_{1} | |||

(P) | =
(P) ⋅ f_{2}(P)
f_{t}(P) +
(P) ⋅ f_{1}(P) | ||

= 0 | |||

(since f _{1}(P) = 0,f_{2}(P) = 0 |

But this contradicts the nonzeroness of the partial at P. Hence
Z(f_{1}) ∩ Z(f_{2}) = ∅. 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.