Puiseux series
Theorem (Newton–Puiseux)
K{{x}} =
S
e⩾1 K((x1/e
)) is algebraically closed.
View f (x, y) = f (x)(y) ∈ K[x][y] ⊂ K((x))[y],
meaning we think of y as an algebraic function of x:
e
C −→ C −→ P1
x .
Let n = degy f .
Then in K{{x}}, f (x)(y) has roots π1, · · · , πn
⇝ For each πj =
P
n⩾n0
anxn/e
,
local parametrisation x = te
, y =
P
n⩾n0
antn
.
This yields all points above x = 0.
For the general case, translate / change variables.
Nicolas Mascot Algebraic curves