[proofplan] We reduce immediately to the affine local computation, because projective local intersection multiplicity is computed in any affine chart containing the point. In the local ring of the affine plane at $p$, the two curve equations vanish at $p$, and their first-order parts define the two tangent lines. Distinct tangent lines mean that these two first-order parts form a basis of the cotangent space $\mathfrak m_p/\mathfrak m_p^2$; [Nakayama's lemma](/theorems/2935) then forces the two equations to generate the maximal ideal. The intersection multiplicity is therefore the dimension of the residue field quotient, which is $1$ over the algebraically closed base field. [/proofplan] [step:Reduce the projective case to an affine chart containing $p$] If $C$ and $D$ are affine curves, set $U=\mathbb A_k^2$. If $C$ and $D$ are projective curves, choose an affine chart $U\subset \mathbb P_k^2$ containing $p$ and identify $U$ with $\mathbb A_k^2$ by its standard affine coordinates. By the definition of local projective intersection multiplicity, the number $I_p(C,D)$ is computed from the corresponding affine curve germs in this chart. The nonsingularity and distinct-tangent hypotheses are also local at $p$, so it remains to prove the affine case. Let $R:=\mathcal O_{\mathbb A_k^2,p}$ be the local ring of the affine plane at $p$, and let $\mathfrak m:=\mathfrak m_p\subset R$ be its maximal ideal. Since $k$ is algebraically closed, the residue field $R/\mathfrak m$ is canonically isomorphic to $k$. [/step] [step:Represent the two curve germs by local equations] Choose affine coordinates on $\mathbb A_k^2$ and let $F,G\in k[x,y]$ be equations for $C$ and $D$ near $p$, respectively. Let $f,g\in R$ denote their images in the local ring $R$. Since $p\in C\cap D$, both functions vanish at $p$, so \begin{align*} f\in \mathfrak m \quad \text{and} \quad g\in \mathfrak m. \end{align*} Set \begin{align*} J:=(f,g)\subset R. \end{align*} The assumption that $C$ and $D$ have no common component through $p$ ensures that the local quotient $R/J$ has finite length, and by the definition of local intersection multiplicity for plane curves, \begin{align*} I_p(C,D)=\dim_k R/J. \end{align*} [/step] [step:Use the distinct tangent lines to span the cotangent space] Let $\bar f$ and $\bar g$ denote the images of $f$ and $g$ in the two-dimensional $k$-[vector space](/page/Vector%20Space) $\mathfrak m/\mathfrak m^2$. Because $C$ is nonsingular at $p$, the linear part $\bar f$ is nonzero; because $D$ is nonsingular at $p$, the linear part $\bar g$ is nonzero. The tangent line to $C$ at $p$ is the kernel of $\bar f$ on $T_p\mathbb A_k^2$, and the tangent line to $D$ at $p$ is the kernel of $\bar g$ on $T_p\mathbb A_k^2$. The two tangent lines are distinct by hypothesis, so $\bar f$ and $\bar g$ are not scalar multiples in $\mathfrak m/\mathfrak m^2$. Since $\mathfrak m/\mathfrak m^2$ is a two-dimensional $k$-vector space, it follows that $\bar f$ and $\bar g$ form a $k$-basis. Equivalently, \begin{align*} \mathfrak m=J+\mathfrak m^2. \end{align*} [guided] The point of passing to $\mathfrak m/\mathfrak m^2$ is that this quotient records exactly the first-order parts of functions vanishing at $p$. The element $\bar f\in \mathfrak m/\mathfrak m^2$ is the linear part of the local equation $f$, and similarly $\bar g$ is the linear part of $g$. For a nonsingular plane curve, the first-order part of a local defining equation is nonzero. Thus nonsingularity of $C$ at $p$ gives \begin{align*} \bar f\ne 0 \quad \text{in} \quad \mathfrak m/\mathfrak m^2, \end{align*} and nonsingularity of $D$ at $p$ gives \begin{align*} \bar g\ne 0 \quad \text{in} \quad \mathfrak m/\mathfrak m^2. \end{align*} The tangent line to $C$ is defined by the vanishing of this linear form on the tangent space $T_p\mathbb A_k^2$, and the tangent line to $D$ is defined by the vanishing of $\bar g$. If $\bar f$ and $\bar g$ were scalar multiples, their kernels would be the same line. Since the tangent lines are distinct, $\bar f$ and $\bar g$ are not scalar multiples. Now $\mathfrak m/\mathfrak m^2$ is the cotangent space of the smooth surface $\mathbb A_k^2$ at $p$, hence it has dimension $2$ over $k$. Two nonzero non-proportional vectors in a two-dimensional vector space form a basis. Therefore the images of $f$ and $g$ span $\mathfrak m/\mathfrak m^2$, which is exactly the statement \begin{align*} \mathfrak m=J+\mathfrak m^2. \end{align*} [/guided] [/step] [step:Apply Nakayama's lemma to identify the generated ideal] Consider the finitely generated $R$-module $\mathfrak m/J$. From \begin{align*} \mathfrak m=J+\mathfrak m^2 \end{align*} we obtain \begin{align*} \mathfrak m/J=\mathfrak m(\mathfrak m/J). \end{align*} By Nakayama's lemma (citing a result not yet in the wiki: Nakayama's Lemma), applied to the finitely generated module $\mathfrak m/J$ over the local ring $(R,\mathfrak m)$, we get \begin{align*} \mathfrak m/J=0. \end{align*} Hence \begin{align*} J=\mathfrak m. \end{align*} [/step] [step:Compute the local quotient and conclude multiplicity one] Since $J=\mathfrak m$, the local quotient is the residue field: \begin{align*} R/J=R/\mathfrak m\cong k. \end{align*} Taking $k$-dimensions gives \begin{align*} I_p(C,D)=\dim_k R/J=\dim_k k=1. \end{align*} This proves the affine case, and the affine-chart reduction proves the projective case as well. [/step]