[proofplan] The pencil of lines through the singular point $p$ is a copy of $\mathbb P_k^1$, and projection from $p$ sends a general point of $C$ to the unique line of the pencil containing it. Since $p$ is a node or cusp, a general line through $p$ has intersection multiplicity exactly $2$ with the cubic at $p$. [Bézout's theorem](/theorems/2161) for plane curves then leaves one residual intersection point on such a line, giving a rational inverse from a dense open subset of the pencil back to $C$. The converse follows from the definition of birationality: rational inverse maps induce inverse pullback homomorphisms on function fields. [/proofplan] [step:Define projection from the singular point as a rational map] Let $\Lambda_p$ denote the set of projective lines $L\subset \mathbb P_k^2$ with $p\in L$. This pencil is naturally a projective line, so we fix an identification $\Lambda_p\cong \mathbb P_k^1$. Define the rational map \begin{align*} \pi_p:C\dashrightarrow \Lambda_p \end{align*} on $C\setminus\{p\}$ by \begin{align*} \pi_p(q)=\overline{pq}, \end{align*} where $\overline{pq}$ is the unique projective line through the distinct points $p$ and $q$. This is a rational map because it is given on the dense open subset $C\setminus\{p\}$ of the irreducible curve $C$. [guided] The projection is not a morphism on all of $C$, because the instruction “send $q$ to the line through $p$ and $q$” is undefined when $q=p$. This is exactly why the theorem states that projection from $p$ defines a rational map. Let $\Lambda_p$ be the pencil of all projective lines through $p$. A pencil of lines through a fixed point in $\mathbb P_k^2$ is parametrised by its directions, hence is isomorphic to $\mathbb P_k^1$. On the dense open subset $C\setminus\{p\}$, every point $q$ determines a unique line $\overline{pq}$ through $p$ and $q$. Thus we obtain a map \begin{align*} \pi_p:C\setminus\{p\}\to \Lambda_p \end{align*} given by \begin{align*} q\mapsto \overline{pq}. \end{align*} Since $C$ is irreducible and $p$ is a closed point, $C\setminus\{p\}$ is dense in $C$. A map defined on a dense open subset is precisely the data of a rational map, so this defines \begin{align*} \pi_p:C\dashrightarrow \Lambda_p\cong \mathbb P_k^1. \end{align*} [/guided] [/step] [step:Use the node or cusp to control intersections with a general line] Because $p$ is a node or a cusp of the plane curve $C$, the multiplicity of $C$ at $p$ is $2$. Equivalently, there is a finite set $E\subset \Lambda_p$ of tangent directions such that, for every line $L\in \Lambda_p\setminus E$, the local intersection multiplicity of $C$ and $L$ at $p$ is \begin{align*} I_p(C,L)=2. \end{align*} Since $C$ is irreducible of degree $3$, no line $L\in\Lambda_p$ is an irreducible component of $C$. We use the standard Bézout theorem for plane curves: if a line $L$ is not a component of a plane cubic $C$, then the sum of local intersection multiplicities of $C$ and $L$ over all intersection points is $3$. Applying this to any $L\in\Lambda_p\setminus E$, we get \begin{align*} \sum_{x\in C\cap L} I_x(C,L)=3. \end{align*} The contribution at $p$ is $2$, so the total intersection multiplicity away from $p$ is $1$: \begin{align*} \sum_{x\in (C\cap L)\setminus\{p\}} I_x(C,L)=1. \end{align*} Therefore, for every $L\in\Lambda_p\setminus E$, there is a unique point $r(L)\in C\setminus\{p\}$ such that \begin{align*} C\cap L=2p+r(L) \end{align*} as an intersection cycle. [guided] The singularity hypothesis is used only through the multiplicity of $C$ at $p$. A node and a cusp are double points, so the multiplicity of the cubic at $p$ is $2$. For a line through $p$ whose direction is not tangent to the singularity, this means the line meets $C$ at $p$ with intersection multiplicity exactly $2$: \begin{align*} I_p(C,L)=2. \end{align*} The tangent directions form a finite subset $E\subset \Lambda_p$: a node has two tangent directions and a cusp has one tangent direction, counted with the usual tangent-cone convention. We remove these finitely many exceptional lines. Now take $L\in\Lambda_p\setminus E$. Since $C$ is irreducible of degree $3$, it cannot contain the line $L$ as a component; otherwise $C$ would be reducible. Thus Bézout's theorem for plane curves applies to the line $L$ and the cubic $C$. It says that the total intersection multiplicity is the product of the degrees: \begin{align*} \sum_{x\in C\cap L} I_x(C,L)=\deg(C)\deg(L)=3\cdot 1=3. \end{align*} We have already accounted for multiplicity $2$ at $p$. Subtracting that contribution gives \begin{align*} \sum_{x\in (C\cap L)\setminus\{p\}} I_x(C,L)=3-2=1. \end{align*} A sum of positive local intersection multiplicities equal to $1$ can have only one term, and that term has multiplicity $1$. Hence there is a unique residual point $r(L)\in C\setminus\{p\}$ on the line $L$, and the intersection cycle has the form \begin{align*} C\cap L=2p+r(L). \end{align*} This residual point is the candidate inverse image of $L$ under the desired inverse rational map. [/guided] [/step] [step:Construct the residual point map as the rational inverse] Let \begin{align*} U:=\Lambda_p\setminus E. \end{align*} Since $E$ is finite and $\Lambda_p\cong \mathbb P_k^1$ is irreducible, $U$ is a dense open subset of $\Lambda_p$. Define \begin{align*} \rho:U\to C \end{align*} by sending $L\in U$ to the residual point $r(L)$ found above. The construction of $r(L)$ is algebraic in the coefficients defining the line $L$: after restricting the homogeneous cubic equation of $C$ to the line $L$, the resulting degree-$3$ binary form has a double zero at $p$, and the remaining linear factor gives $r(L)$. Hence $\rho$ is a regular map on $U$, and therefore determines a rational map \begin{align*} \rho:\Lambda_p\dashrightarrow C. \end{align*} For every $L\in U$, the point $\rho(L)$ lies on $L$ and is distinct from $p$, so \begin{align*} \pi_p(\rho(L))=L. \end{align*} Thus $\pi_p\circ \rho=\operatorname{id}_U$. [/step] [step:Show the residual point map also inverts projection on a dense open subset of the cubic] Let \begin{align*} V:=\rho(U)\subset C. \end{align*} For each $L\in U$, the residual point $\rho(L)$ is the unique point of $(C\cap L)\setminus\{p\}$. Conversely, if $q\in V$, then $q=\rho(L)$ for some $L\in U$, and the line through $p$ and $q$ is $L$. Hence \begin{align*} \rho(\pi_p(q))=q \end{align*} for every $q\in V$. The set $V$ is dense in $C$ because $\rho$ is inverse to $\pi_p$ on $U$ and $\pi_p$ is dominant onto the pencil: a general line through $p$ has a residual point on $C$. Therefore $\pi_p$ and $\rho$ restrict to inverse isomorphisms between dense open subsets of $C$ and $\Lambda_p$. This proves that \begin{align*} \pi_p:C\dashrightarrow \Lambda_p\cong \mathbb P_k^1 \end{align*} is birational. [/step] [step:Identify function fields from a birational parametrisation] For the converse, let $\varphi:\mathbb P_k^1\dashrightarrow C$ be a birational rational parametrisation. By hypothesis, there exist dense open subsets $U\subset \mathbb P_k^1$ and $V\subset C$ such that $\varphi$ restricts to an isomorphism \begin{align*} \varphi|_U:U\to V. \end{align*} Let \begin{align*} \psi:V\to U \end{align*} denote its inverse regular map. Pullback of rational functions along $\varphi$ gives a field homomorphism \begin{align*} \varphi^*:k(C)\to k(\mathbb P_k^1). \end{align*} Pullback along $\psi$ gives a field homomorphism \begin{align*} \psi^*:k(\mathbb P_k^1)\to k(C). \end{align*} Because $\psi\circ\varphi=\operatorname{id}_U$ and $\varphi\circ\psi=\operatorname{id}_V$, these pullback maps satisfy \begin{align*} \varphi^*\circ \psi^*=\operatorname{id}_{k(\mathbb P_k^1)} \end{align*} and \begin{align*} \psi^*\circ \varphi^*=\operatorname{id}_{k(C)}. \end{align*} Thus $\varphi^*$ is an isomorphism of fields. Finally, the function field of $\mathbb P_k^1$ is the rational function field in one variable: \begin{align*} k(\mathbb P_k^1)=k(t). \end{align*} Therefore \begin{align*} k(C)\cong k(t). \end{align*} This completes the proof. [/step]