$I(P, C \cap D)=1$ iff $P$ is nonsingular on both $C$ and $D$ and the tangent lines to $C$ and $D$ at $P$ are distinct. While reading Milne's elliptic curve book, I am stuck at one more point that if $I(P, C \cap D)=1$ iff $P$ is nonsingular on both $C$ and $D$ and the tangent lines to $C$ and $D$ at $P$ are distinct. Here $C,D$ represents curves corresponding to polynomials say $f,g \in k[X,Y]$.
Now let's try the $\Leftarrow$ part,
WLOG we take $P=(0,0)$ Now we can write $f=f_1+...+f_r$ & $g=g_1+...+g_m$ where $f_i,g_i$'s are homogeneous polynomial in $X,Y$ of degree $i$. Now $P$ is nonsingular that implies $f_1(P)g_1(P) \neq 0$ and $f_1=aX+bY$ and $g_1=cX+dY$ are distinct, so $I(P, C \cap D)=I(f,g)$. Now is it $=I(f_1,g_1)?$ if it is so then $I(aX+bY,cX+dY)=1$ As $(0,0)$ is in the intersection.
Conversely, $\Rightarrow$ part I am not getting, as well.
Please help.
Moreover, please don't use the general result like $I(p, C\cap D) \geq m_p(C)m_P(D)$.
 A: Let's try to give an attempt to your question 
$\Leftarrow$ part
Let $f$ and $g$ are two fuction where $f_1$ and $g_1$ are non vanishing.
Now by resultant $\exists a(X,Y),b(X,Y)$ s.t $af+bg=r \in k[X]$ and $deg_Y(b)<deg_Y(f)$, $deg_Y(a)<deg_Y(g)$.
If $deg_Y(f)<deg_Y(g)$, write 
$I(f,g)=I(f,bg)-I(f,b)=I(f,r)-I(f,b)$ 
Now cosider $deg_Y(b)<deg_Y(f)$ 
then $I(f,g)=I(f,bg)-I(f,b)=I(f,r)-I(f,b)=I(f,r)-I(af,b)+I(a,b)=I(f,r)-I(r-bg,b)+I(a,b)=I(f,r)-I(r,b)+I(a,b)$
Then by resultant theory $\exists a_1(X,Y),b_1(X,Y)$ s.t $a_1a+b_1b=r_1\in k[X]$ s.t $deg_Y(b_1)<deg_Y(a)$ and $deg_Y(a_1)<deg_Y(b)$
If $deg_Y(a)<deg_Y(b)$ then $I(a,b)=I(a,r_1)-I(r_1,b)+I(a_1,b_1)$
then $I(f,g)=I(f,r)-I(r,b)+I(a,b)=I(f,r)-I(r,b)+I(a,r_1)-I(r_1,b)+I(a_1,b_1)$
Now as the $f_1$ and $g_1$ are non zero and distinct we get $I(f,r)=I(r,b)=I(a,r_1)=I(r_1,b)=...=0$
Continue in this fashion untill $Y$ will be eliminated from one function, say, from $g$. Then $g(X)=Xg_0(X)$ s.t $g_0(0) \neq 0$ [as the $f_1$ and $g_1$ are non zero and distinct],
so $I(f,g)=I(f,X)$ and after subtracting a multiple of $X$ from $f(X,Y)$ we can assume that it is a polynomial in $Y$ alone [as the $f_1$ and $g_1$ are non zero and distinct], 
write $f(Y)=Yf_0(Y)$ where $f_0(0) \neq 0$. Then $I(f,g)=I(f,X)=I(Y,X)=1$
$\Rightarrow$ part
Conversely, $I(f,g)=\dim_k k[X,Y]_{(0,0)}/<f,g>=1=\dim_k k[X,Y]_{(0,0)}/<X,Y>$, we have $<f,g>=<X,Y>$ hence $P$ is nosingular on both $C$ and $D$ and the tangent lines there at $P(0,0)$ are distinct.
Please check, whether I have made any mistakes or not.
