[proofplan] The proof splits on the possible degrees of $C$ and $D$. If either has degree $\leq 1$, then [Bézout for Lines](/theorems/2162) directly gives $|C \cap D| \leq \deg C \cdot \deg D \leq 1 \cdot 2 = 2 \leq 4$. The remaining case is $\deg C = \deg D = 2$. Here we further split: if either conic is reducible (a pair of lines or a double line), apply [Bézout for Lines](/theorems/2162) to each component. If both conics are irreducible, parametrise $C$ via [Irreducible Conics Are Smooth](/theorems/2163) by the Veronese-type map $\Phi: \mathbb{P}^1_k \to C$ given by quadratic forms, pull back the equation of $D$, and note that the pullback is a binary form of degree at most $2 \cdot 2 = 4$, whose roots count the intersection points. The algebraic-closure of $k$ then bounds the root count by the degree. [/proofplan]

[step:Reduce to the case $\deg C = \deg D = 2$ by handling lower-degree cases via Bézout] We split on $\min(\deg C, \deg D)$. \textbf{Case A: $\min(\deg C, \deg D) \leq 1$.} Without loss of generality assume $\deg C \leq 1$. If $\deg C = 0$, then $C = V(F)$ for a non-zero constant $F \in k^*$, so $V(F) = \varnothing$ and $C \cap D = \varnothing$; the bound $0 \leq 4$ is trivial. If $\deg C = 1$, then $C$ is a line $L$. By hypothesis $C, D$ share no common component, so $L \not\subset D$ (a curve of degree $\leq 2$ has line components only if reducible, and a shared line would be a shared component). Apply [Bézout for Lines](/theorems/2162) to $L$ and $D$: this requires $L \not\subset D$, which we just verified, and gives \begin{align*} |L \cap D| \leq \deg D \leq 2 \leq 4. \end{align*} \textbf{Case B: $\deg C = \deg D = 2$.} This is the substantive case, treated in the following steps. [/step]

[step:Reduce Case B further by splitting on irreducibility of $C$ and $D$] Within Case B, we split on whether each of $C, D$ is irreducible. A conic $V(F)$ with $\deg F = 2$ is either irreducible, or factors as $F = L_1 L_2$ with $L_1, L_2$ linear (possibly equal, in which case $V(F)$ is a "double line"). We use this dichotomy. \textbf{Subcase B1: At least one of $C, D$ is reducible.} Without loss of generality assume $D = V(L_1 L_2)$ with $L_1, L_2$ linear forms (possibly equal). Then $D = L_1 \cup L_2$ as sets in $\mathbb{P}^2_k$ where $L_i := V(L_i)$. We have \begin{align*} C \cap D = (C \cap L_1) \cup (C \cap L_2). \end{align*} For each $i = 1, 2$: since $C$ and $D$ share no common component, the line $L_i$ is not a component of $C$. If $C$ is irreducible of degree $2$, then $L_i$ being a component of $C$ would force $C = L_i$ as varieties, but $C$ has degree $2$ and $L_i$ has degree $1$, contradiction; alternatively, if $C$ is reducible with linear components $M_1, M_2$, then $L_i$ being a common component would mean $L_i \in \{M_1, M_2\}$, again contradicting "no common component". In either case $L_i \not\subset C$. By [Bézout for Lines](/theorems/2162) applied to $L_i$ and $C$ (with $L_i \not\subset C$), we have $|L_i \cap C| \leq \deg C = 2$. Hence \begin{align*} |C \cap D| \leq |C \cap L_1| + |C \cap L_2| \leq 2 + 2 = 4. \end{align*} \textbf{Subcase B2: Both $C$ and $D$ are irreducible.} Treated in the next steps. [/step]

[step:Parametrise $C$ via the smooth-conic isomorphism] Assume Subcase B2: $C, D$ both irreducible of degree $2$. By [Irreducible Conics Are Smooth](/theorems/2163), there is an isomorphism \begin{align*} \Phi: \mathbb{P}^1_k &\to C \\ [Y_0 : Y_1] &\mapsto [Q_0(Y_0, Y_1) : Q_1(Y_0, Y_1) : Q_2(Y_0, Y_1)], \end{align*} where $Q_0, Q_1, Q_2 \in k[Y_0, Y_1]$ are quadratic forms (homogeneous of degree $2$), and where the hypothesis $\operatorname{char} k \neq 2$ in the cited theorem is met by our standing assumption. Since $\Phi$ is an isomorphism of varieties, it is in particular a bijection on points. Let $G \in k[X_0, X_1, X_2]$ be a homogeneous polynomial of degree $2$ with $D = V(G)$. [/step]

[step:Pull back the equation of $D$ along $\Phi$ to obtain a binary form of degree at most $4$] Define \begin{align*} H: k^2 &\to k \\ (Y_0, Y_1) &\mapsto G(Q_0(Y_0, Y_1), Q_1(Y_0, Y_1), Q_2(Y_0, Y_1)). \end{align*} Since $G$ is homogeneous of degree $2$ in $(X_0, X_1, X_2)$, and each $Q_i$ is homogeneous of degree $2$ in $(Y_0, Y_1)$, the composition $H$ is a homogeneous polynomial in $(Y_0, Y_1)$ of degree $2 \cdot 2 = 4$: \begin{align*} H \in k[Y_0, Y_1], \quad \deg H \leq 4 \text{ (and homogeneous of pure degree } 4\text{ when non-zero)}. \end{align*} We claim $H$ is non-zero as a polynomial. Suppose for contradiction $H \equiv 0$. Then $G(Q_0, Q_1, Q_2) = 0$ as a polynomial in $k[Y_0, Y_1]$, which means $G$ vanishes identically on the image $\Phi(\mathbb{P}^1_k) = C$. By [Hilbert's Nullstellensatz](/theorems/2124) applied to the projective coordinate ring (specifically, the homogeneous Nullstellensatz: if a homogeneous polynomial $G$ vanishes on $V(F)$ with $F$ homogeneous and $V(F) \neq \varnothing$, then some power $G^N \in (F)$, and since both $F$ and $G$ have the same degree $2$ and are irreducible, this forces $G$ to be a scalar multiple of $F$), $G$ would be a $k^*$-multiple of $F$ (the defining equation of $C$). But this means $D = V(G) = V(F) = C$, so $C$ and $D$ share the irreducible component $C$, contradicting the hypothesis "no common component". Hence $H \neq 0$. A direct argument avoiding the Nullstellensatz: since $\Phi$ is a bijection $\mathbb{P}^1_k \to C$, $H$ vanishes identically on $\mathbb{P}^1_k$ iff $G$ vanishes identically on $C$ iff $C \subseteq V(G) = D$. Since $C$ is irreducible, $C \subseteq D$ would make $C$ an irreducible component of $D$ (as $C$ is closed and irreducible and $\dim C = \dim D = 1$); since $D$ has degree $2$ and $C$ has degree $2$, this would force $D = C$, again contradicting no common component. [/step]

[step:Identify intersection points of $C \cap D$ with zeros of $H$ on $\mathbb{P}^1_k$] For $[Y_0 : Y_1] \in \mathbb{P}^1_k$, the image $\Phi([Y_0 : Y_1]) \in C$ lies in $D = V(G)$ if and only if $G(Q_0, Q_1, Q_2)(Y_0, Y_1) = H(Y_0, Y_1) = 0$. Hence \begin{align*} \Phi^{-1}(C \cap D) = \{[Y_0 : Y_1] \in \mathbb{P}^1_k : H(Y_0, Y_1) = 0\} = V_{\mathbb{P}^1_k}(H). \end{align*} Since $\Phi: \mathbb{P}^1_k \to C$ is a bijection, $|C \cap D| = |V_{\mathbb{P}^1_k}(H)|$. [/step]

[step:Bound the number of zeros of a non-zero binary form of degree $4$ over an algebraically closed field]By Step 4, $H \in k[Y_0, Y_1]$ is a non-zero homogeneous polynomial of degree exactly $4$ (homogeneity is preserved under composition of homogeneous maps). Over an algebraically closed field $k$, a non-zero binary form of degree $d$ factors completely into linear factors: \begin{align*} H(Y_0, Y_1) = c \prod_{s=1}^{r} (\beta_s Y_0 - \alpha_s Y_1)^{m_s}, \end{align*} with $c \in k^*$, distinct points $[\alpha_s : \beta_s] \in \mathbb{P}^1_k$, and multiplicities $m_s \geq 1$ summing to $d = 4$. The zero set $V_{\mathbb{P}^1_k}(H) = \{[\alpha_s : \beta_s] : 1 \leq s \leq r\}$ has cardinality $r$, and \begin{align*} r \leq \sum_{s=1}^{r} m_s = 4. \end{align*} Hence $|V_{\mathbb{P}^1_k}(H)| \leq 4$, and combined with Step 5, \begin{align*} |C \cap D| = |V_{\mathbb{P}^1_k}(H)| \leq 4. \end{align*}[/step]

[guided]Why does the binary form factor into $4$ linear factors? Apply the [Fundamental Theorem of Algebra](/theorems/???) to the dehomogenisation. Pick any chart of $\mathbb{P}^1_k$ where the leading coefficient of $H$ does not vanish — at least one such chart exists because $H$ is non-zero. Set $Y_1 = 1$, say, to obtain $\tilde H(t) := H(t, 1) \in k[t]$ of degree $\leq 4$. Over $k$ algebraically closed, $\tilde H$ factors as $c \prod (t - \alpha_s)^{m_s}$ with $\sum m_s = \deg \tilde H$. Re-homogenising of $\tilde H$ gives a factorisation of $H$, with possibly an additional factor of $Y_1^{4 - \deg \tilde H}$ representing the "zero at infinity" $[1:0]$. The number of distinct zeros is at most the degree because each distinct zero contributes at least multiplicity $1$, and the total multiplicity is exactly the degree. The bound $4 = 2 \cdot 2$ is the special case of [Bézout's Theorem](/theorems/???) (the full version, for general plane curves) with both degrees equal to $2$. The argument here proves this special case directly, using only Bézout for lines and the rational parametrisation of smooth conics — both of which are elementary, while the full Bézout requires the theory of intersection multiplicities and the fact that local intersection algebras have finite length.[/guided]

[step:Combine all cases to conclude] We have shown: \begin{itemize} \item In Case A ($\min \deg \leq 1$): $|C \cap D| \leq 2 \leq 4$. \item In Subcase B1 (one conic reducible): $|C \cap D| \leq 4$. \item In Subcase B2 (both conics irreducible): $|C \cap D| \leq 4$. \end{itemize} These cases exhaust all possibilities for plane curves $C, D$ of degree at most $2$ sharing no component, so $|C \cap D| \leq 4$ in all cases. This completes the proof. [/step]