[proofplan]
We translate coordinates so that $p=(0,0)$ and work in the regular local ring $\mathcal O_{\mathbb A_k^2,p}$ with maximal ideal generated by the coordinate functions. Nonsingularity says that the linear parts of $F$ and $G$ are nonzero, while distinct tangent lines say that those linear parts are linearly independent in $\mathfrak m/\mathfrak m^2$. Hence the ideal generated by $F$ and $G$ maps onto $\mathfrak m/\mathfrak m^2$, so [Nakayama's lemma](/theorems/2935) gives $(F,G)=\mathfrak m$. The quotient is therefore the residue field $k$, whose $k$-dimension is $1$.
[/proofplan]
[step:Move the point to the origin and set up the local ring]
Write $p=(a,b)\in k^2$. The affine change of coordinates
\begin{align*}
x' = x-a
\end{align*}
and
\begin{align*}
y' = y-b
\end{align*}
induces an isomorphism of local rings at $p$ and at $(0,0)$, and carries the ideal generated by $F$ and $G$ to the ideal generated by the translated polynomials. Since local intersection multiplicity is the $k$-dimension of the corresponding local quotient, this change of coordinates preserves $I_p(F,G)$.
Thus we may assume $p=(0,0)$. Set
\begin{align*}
R:=k[x,y]
\end{align*}
and let
\begin{align*}
\mathfrak q:=(x,y)\trianglelefteq R.
\end{align*}
Define the local ring
\begin{align*}
A:=R_{\mathfrak q}
\end{align*}
and its maximal ideal
\begin{align*}
\mathfrak m:=\mathfrak q A.
\end{align*}
Because $p$ is a $k$-rational point, the residue field $A/\mathfrak m$ is canonically isomorphic to $k$.
[/step]
[step:Identify the linear parts with elements of $\mathfrak m/\mathfrak m^2$]
Since $F(0,0)=G(0,0)=0$, both $F$ and $G$ lie in $\mathfrak m$. Write $F_1\in k[x,y]_1$ for the degree-one homogeneous part of $F$, and write $G_1\in k[x,y]_1$ for the degree-one homogeneous part of $G$. The images of $F$ and $G$ in the two-dimensional $k$-[vector space](/page/Vector%20Space) $\mathfrak m/\mathfrak m^2$ are exactly the classes of $F_1$ and $G_1$.
The nonsingularity of $C=V(F)$ at $(0,0)$ means that $F_1\neq 0$, and the nonsingularity of $D=V(G)$ at $(0,0)$ means that $G_1\neq 0$. The tangent line to $C$ at $(0,0)$ is defined by $F_1=0$, and the tangent line to $D$ at $(0,0)$ is defined by $G_1=0$. Since these tangent lines are distinct, the nonzero linear forms $F_1$ and $G_1$ are not scalar multiples of one another. Therefore their classes form a $k$-basis of $\mathfrak m/\mathfrak m^2$.
[/step]
[step:Lift generation modulo $\mathfrak m^2$ to generation of $\mathfrak m$]
Let
\begin{align*}
J:=(F,G)A\trianglelefteq A.
\end{align*}
From the previous step, the image of $J$ in $\mathfrak m/\mathfrak m^2$ is all of $\mathfrak m/\mathfrak m^2$. Equivalently,
\begin{align*}
\mathfrak m=J+\mathfrak m^2.
\end{align*}
Apply Nakayama's lemma to the finitely generated $A$-module $\mathfrak m$ and its submodule $J\subseteq \mathfrak m$. Since $A$ is local with maximal ideal $\mathfrak m$, the equality $\mathfrak m=J+\mathfrak m\mathfrak m$ implies
\begin{align*}
\mathfrak m=J.
\end{align*}
[guided]
The point of passing to $\mathfrak m/\mathfrak m^2$ is that this quotient records the first-order part of functions vanishing at the point. We have already proved that the classes of $F$ and $G$ span this quotient. In module language, this says that the submodule $J=(F,G)A$ of $\mathfrak m$ has the same image as $\mathfrak m$ after reducing modulo $\mathfrak m^2$.
That statement is precisely
\begin{align*}
\mathfrak m=J+\mathfrak m^2.
\end{align*}
Since $\mathfrak m^2=\mathfrak m\mathfrak m$, this can be rewritten as
\begin{align*}
\mathfrak m=J+\mathfrak m\mathfrak m.
\end{align*}
Now we invoke Nakayama's lemma in the following standard form: if $(A,\mathfrak m)$ is a local ring, $M$ is a finitely generated $A$-module, and $N\subseteq M$ is a submodule such that $M=N+\mathfrak m M$, then $M=N$. Here $M=\mathfrak m$ and $N=J$. The module $\mathfrak m$ is finitely generated because $\mathfrak m=(x,y)A$. Therefore Nakayama's lemma applies and gives
\begin{align*}
\mathfrak m=J.
\end{align*}
This is the algebraic form of the geometric statement that two independent tangent directions cut out the point to first order with no higher multiplicity.
[/guided]
[/step]
[step:Compute the local quotient and conclude the multiplicity is one]
By the definition of local intersection multiplicity used in the statement,
\begin{align*}
I_p(F,G)=\dim_k A/J.
\end{align*}
The previous step gives $J=\mathfrak m$, so
\begin{align*}
A/J=A/\mathfrak m.
\end{align*}
Because $p=(0,0)$ is a $k$-rational point of $\mathbb A_k^2$, the residue field $A/\mathfrak m$ is $k$. Hence
\begin{align*}
I_p(F,G)=\dim_k k=1.
\end{align*}
This proves the claimed multiplicity-one result.
[/step]
