I.5.3

6/15/2021

So we can assume without loss of generality that P = (0, 0) and thus I'll write
μ = μ_{P}.

Let's write f_{1} = ax + by. From 5.1, we know that P is singular iff

(P) | = 0 | ||

& (P) | = 0 | ||

⇐⇒ | |||

(P) + (P) + (P) + + (P) | = 0 | ||

& (P) + (P) + (P) + + (P) | = 0 | ||

⇐⇒ | |||

0 + (P) + 0 + + 0 | = 0 | ||

&0 + (P) + 0 + + 0 | = 0 | ||

⇐⇒ | |||

(P) | = 0 | ||

& (P) | = 0 | ||

⇐⇒ | |||

a | = 0 | ||

&b | = 0 | ||

⇐⇒ | |||

f_{1} | = 0 |

(Those 0s come from the fact that f

I.e. we showed that

P singular | ⇐⇒f_{1} = 0 |

In other words,

P nonsingular | ⇐⇒f_{1}≠0 | ||

⇐⇒μ(Y ) ≤ 1 |

As the exercise states, the condition P ∈ Y is equivalent to μ(Y ) > 0⇐⇒μ(Y ) ≥ 1. So

P nonsingular and P ∈ Y | ⇐⇒μ(Y ) = 1 | ||

Done.

Here's 5.1. So, ASSUMING THAT THE ONLY SINGULAR POINTS ARE THE
ORIGIN (LOL), all we have to do is look at each equation and take the degree of
the minimum degree monomial

x^{4} + y^{4} - x^{2} | μ = 2 | ||

x^{6} + y^{6} - xy | μ = 2 | ||

y^{2} + x^{4} + y^{4} - x^{3} | μ = 2 | ||

x^{4} + y^{4} - x^{2}y - xy^{2} | μ = 3 |

Done.