TITLE: Non singular point of a projective plane curve
QUESTION [0 upvotes]: In problem 5.1 of Fulton's Algebraic curves, we're asked to show that a point $P\in\mathbb P^2$, $P=[P_1:P_2:P_3]$ is multiple iff $F(P)=F_X(P)=F_Y(P)=F_Z(P)=0$.
Here $P$ is said to be multiple if $m_P(F):=\text{dim}_K(n_P(F)^n/n_P(F)^{n+1})>1$ where $n_p(F)\subseteq O_p(V(F))$ is the maximal ideal, with $V(F)$ a projective variety. We also have that if $P=[P_1,P_2,1]$, $m_P(F)=m_{(P_1,P_2)}(F_*)$ which is the characterization I'm using.
Here are my thoughts on each direction.
Suppose we have the equalities, and suppose that $P\in U_3$ (it can be in others but it doesn't change my reasoning) and $P_3=1$ up to division by $P_3$. We then have $F_X(P)=F_Y(P)=0 \iff F_X(P_1,P_2,1)=F_Y(P_1,P_2,1)\iff (F_*)_X(P_1,P_2)=(F_*)_Y(P_1,P_2)=0$ so $(P_1,P_2)$ is singular on $F_*$ so $m_P(F_*)>1$ and so $P$ is singular on $F$.
Is it ok ?
Now if $P$ is multiple, we have that $m_P(F)>1$ so $m_P(F_*)>1$ so we get $(F_*)_X(P_1,P_2)=(F_*)_Y(P_1,P_2)=0 \implies F_X(P)=F_Y(P)=0$.
From here I guess we have to use that $F$ is homogeneous  to deduce that $F_Z(P)=0$ but I'm struggling.

REPLY [0 votes]: If we know that for an homogeneous polynomial in $K[x_1,\dots,x_n]$ of degree $d$, we have that $d.F=\sum_{i=1}x_iF_{x_i}$, we can use it and it becomes quite easy :
$$3.F=XF_X+YF_Y+ZF_Z\implies 0=P_3F_Z(P)\iff F_Z(P)=0$$ since we supposed $P_3=1$, and we used $F(P)=F_X(P)=F_Y(P)=0$
{"set_name": "stack_exchange", "score": 0, "question_id": 4456157}