I.5.3
6/15/2021
So we can assume without loss of generality that P = (0, 0) and thus I'll write
μ = μP.
Let's write f1 = ax + by. From 5.1, we know that P is singular iff
![]() | = 0 | ||
&
![]() | = 0 | ||
⇐⇒ | |||
![]() ![]() ![]() ![]() ![]() | = 0 | ||
&
![]() ![]() ![]() ![]() ![]() | = 0 | ||
⇐⇒ | |||
0 +
![]() ![]() | = 0 | ||
&0 +
![]() ![]() | = 0 | ||
⇐⇒ | |||
![]() | = 0 | ||
&
![]() | = 0 | ||
⇐⇒ | |||
a | = 0 | ||
&b | = 0 | ||
⇐⇒ | |||
f1 | = 0 |
P singular | ⇐⇒f1 = 0 |
P nonsingular | ⇐⇒f1≠0 | ||
⇐⇒μ(Y ) ≤ 1 |
P nonsingular and P ∈ Y | ⇐⇒μ(Y ) = 1 | ||
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
x4 + y4 - x2 | μ = 2 | ||
x6 + y6 - xy | μ = 2 | ||
y2 + x4 + y4 - x3 | μ = 2 | ||
x4 + y4 - x2y - xy2 | μ = 3 |