[proofplan] The strategy is to reduce to a univariate polynomial root count. After a projective change of coordinates we may assume $L = V(X_2)$, and after a further change we may assume $C$ does not pass through the point $[0:1:0]$. The line $L$ is then identified with $\mathbb{P}^1_k$ via $[X_0 : X_1] \mapsto [X_0 : X_1 : 0]$, and the equation defining $C \cap L$ becomes the binary form $f(X_0, X_1) := F(X_0, X_1, 0) \in k[X_0, X_1]$. The hypothesis $L \not\subset C$ ensures $f$ is non-zero, and the avoidance of $[0:1:0]$ ensures $\deg f = c$. The intersection points correspond bijectively to the roots of $f$, and since $k$ is algebraically closed every binary form of degree $c$ factors completely into $c$ linear factors counted with multiplicity. The intersection multiplicity at each point is matched with the root multiplicity of the corresponding linear factor. [/proofplan]

[step:Choose projective coordinates so that $L = V(X_2)$ and $[0:1:0]\notin C$] We are free to apply any element of $\mathrm{PGL}_3(k)$ to $\mathbb{P}^2_k$, since this group acts transitively on lines and intersection multiplicities are invariant under projective changes of coordinates. First, since $\mathrm{PGL}_3(k)$ acts transitively on the set of lines in $\mathbb{P}^2_k$, choose a projective change of coordinates carrying $L$ to the line $V(X_2)$. After this change of coordinates we have $L = V(X_2) = \{[X_0 : X_1 : 0] : [X_0 : X_1] \in \mathbb{P}^1_k\}$. Second, we claim there exists a point $q \in L$ with $q \notin C$. Indeed, if $L \subset C$ this would contradict the hypothesis $L \not\subset C$. Hence the set $L \setminus C$ is non-empty. Choose $q \in L \setminus C$. Apply a further projective change of coordinates fixing the line $L = V(X_2)$ pointwise on a single point but moving $q$ to $[0:1:0]$ — concretely, the projective transformations of $\mathbb{P}^2_k$ that fix the line $V(X_2)$ as a set act on it as $\mathrm{PGL}_2(k)$ on $\mathbb{P}^1_k$, which is triply transitive, so such a transformation exists. After this further change of coordinates, $L = V(X_2)$ still holds and $[0:1:0] \notin C$.[/step]

[guided]Why two changes of coordinates rather than one? The first fixes the geometric position of $L$. The second, more subtle move ensures the curve $C$ avoids the special point $[0:1:0]$, which will play the role of the "point at infinity" for our parametrisation of $L$. Without this avoidance step, the binary form representing $C \cap L$ could fail to have full degree $c$, and the root count would undercount. The reduction is valid because intersection multiplicities are intrinsic — they do not depend on the coordinate system. Concretely, $m_p(L, C)$ can be defined as the dimension of the local intersection algebra $\mathcal{O}_{\mathbb{P}^2_k, p} / (F_L, F_C)$ where $F_L, F_C$ are local equations for $L, C$ at $p$; this dimension is invariant under any local isomorphism, including those induced by elements of $\mathrm{PGL}_3(k)$. Likewise the cardinality $|L \cap C|$ is a set-theoretic invariant. To find $q \in L \setminus C$: the hypothesis $L \not\subset C$ means $L \setminus C \neq \varnothing$, since $L \subset C$ is the negation of "there is a point of $L$ not on $C$". So such a $q$ exists. To move $q$ to $[0:1:0]$ while preserving $L$: the subgroup of $\mathrm{PGL}_3(k)$ fixing the line $V(X_2)$ (setwise) acts on this line as the full $\mathrm{PGL}_2(k)$. This action is triply transitive on $\mathbb{P}^1_k$, so in particular it can carry $q$ to any prescribed point on $L$, e.g.\ $[0:1:0]$.[/guided]

[step:Restrict $F$ to the line $L$ and identify the resulting form] With coordinates chosen as in Step 1, define the binary form \begin{align*} f: k^2 &\to k \\ (X_0, X_1) &\mapsto F(X_0, X_1, 0). \end{align*} Then $f \in k[X_0, X_1]$ is a homogeneous polynomial of degree at most $c$. We claim $f$ is non-zero and $\deg f = c$. Non-zero: if $f \equiv 0$, then $F(X_0, X_1, 0) = 0$ as a polynomial, which means the entire line $L = V(X_2)$ is contained in $V(F) = C$, contradicting $L \not\subset C$. Degree exactly $c$: write \begin{align*} F(X_0, X_1, X_2) = \sum_{i + j + l = c} a_{i,j,l}\, X_0^i X_1^j X_2^l. \end{align*} Setting $X_2 = 0$ gives \begin{align*} f(X_0, X_1) = F(X_0, X_1, 0) = \sum_{i + j = c} a_{i,j,0}\, X_0^i X_1^j, \end{align*} which is homogeneous of degree exactly $c$ in $(X_0, X_1)$, provided some $a_{i,j,0}$ with $i + j = c$ is non-zero. The coefficient of $X_1^c$ in $f$ is $a_{0,c,0}$, which equals $F(0, 1, 0)$. Since $[0:1:0] \notin C = V(F)$ by Step 1, $F(0,1,0) \neq 0$, so $a_{0,c,0} \neq 0$ and $f$ has degree exactly $c$, with leading term $a_{0,c,0} X_1^c$. [/step]

[step:Match intersection points of $L$ and $C$ with zeros of $f$ on $\mathbb{P}^1_k$] Identify the line $L$ with $\mathbb{P}^1_k$ via the isomorphism \begin{align*} \iota: \mathbb{P}^1_k &\to L \subset \mathbb{P}^2_k \\ [X_0 : X_1] &\mapsto [X_0 : X_1 : 0]. \end{align*} A point $p = [X_0 : X_1 : 0] \in L$ lies in $C$ if and only if $F(X_0, X_1, 0) = 0$, i.e.\ $f(X_0, X_1) = 0$. Hence under $\iota^{-1}$, \begin{align*} L \cap C \xleftrightarrow{\;1\text{-}1\;} V_{\mathbb{P}^1_k}(f) = \{[X_0 : X_1] \in \mathbb{P}^1_k : f(X_0, X_1) = 0\}. \end{align*} [/step]

[step:Factor $f$ into linear forms over $k$ algebraically closed]The binary form $f \in k[X_0, X_1]$ is homogeneous of degree exactly $c$. By Step 2, the coefficient of the monomial $X_1^c$ in $f$ is $a_{0,c,0} = F(0,1,0) \neq 0$, since $[0:1:0] \notin C$. \textbf{Choose the dehomogenisation chart.} We dehomogenise in the chart $\{X_1 \neq 0\}$ on $\mathbb{P}^1_k$, using the affine coordinate $s := X_0/X_1$. Substituting $X_1 = 1$: \begin{align*} \hat f(s) := f(s, 1) = \sum_{i + j = c} a_{i,j,0}\, s^i \cdot 1^j = \sum_{i=0}^{c} a_{i,\, c - i,\, 0}\, s^i \in k[s]. \end{align*} The coefficient of $s^i$ in $\hat f$ is $a_{i,\, c-i,\, 0}$; in particular: \begin{align*} \hat f(0) &= a_{0, c, 0} \neq 0 \quad (\text{constant term}), \\ \text{coefficient of } s^c \text{ in } \hat f &= a_{c, 0, 0} = F(1, 0, 0) \quad (\text{may or may not vanish}). \end{align*} \textbf{Degree of $\hat f$.} The degree of $\hat f$ is the largest $i$ such that $a_{i,\, c-i,\, 0} \neq 0$. Denote \begin{align*} r' &:= \deg \hat f \in \{0, 1, \ldots, c\}, & m_\infty &:= c - r' \in \{0, 1, \ldots, c\}. \end{align*} Note $\hat f \not\equiv 0$ since its constant term $a_{0,c,0}$ is non-zero, so $r' \geq 0$ is well-defined. \textbf{Factor $\hat f$ over $k$.} Since $k$ is algebraically closed, every non-constant polynomial in $k[s]$ factors into linear factors by the *Fundamental Theorem of Algebra*. If $r' \geq 1$, write \begin{align*} \hat f(s) = a_{r',\, c - r',\, 0} \prod_{s'=1}^{r} (s - \alpha_{s'})^{m_{s'}}, \end{align*} with distinct $\alpha_1, \ldots, \alpha_r \in k$ and multiplicities $m_{s'} \geq 1$ satisfying $\sum_{s'=1}^{r} m_{s'} = r'$. The leading coefficient of $\hat f$ is $a_{r',\, c - r',\, 0}$, the (non-zero) coefficient of $s^{r'}$. Moreover, since $\hat f(0) = a_{0, c, 0} \neq 0$, none of the roots $\alpha_{s'}$ equals $0$. (If $r' = 0$, then $\hat f$ is a nonzero constant $a_{0,c,0}$ and we take $r = 0$ with an empty product.) \textbf{Rehomogenise to obtain a factorisation of $f$.} The homogenisation of $\hat f$ to degree $c$ in the variables $(X_0, X_1)$ is \begin{align*} f(X_0, X_1) = X_1^c \, \hat f(X_0/X_1). \end{align*} This identity holds because each monomial $a_{i,\, c-i,\, 0}\, s^i$ in $\hat f(s)$ rehomogenises to $a_{i,\, c-i,\, 0}\, X_0^i X_1^{c-i}$ (multiplying by $X_1^c$ and replacing $s = X_0/X_1$), which is the $(i, c-i)$-term of $f$. Substituting the factorisation of $\hat f$: \begin{align*} f(X_0, X_1) &= X_1^c \cdot a_{r',\, c - r',\, 0} \prod_{s'=1}^{r} \left(\frac{X_0}{X_1} - \alpha_{s'}\right)^{m_{s'}} \\ &= a_{r',\, c - r',\, 0}\, X_1^{c - r'} \prod_{s'=1}^{r} (X_0 - \alpha_{s'} X_1)^{m_{s'}} \\ &= a_{r',\, c - r',\, 0}\, X_1^{m_\infty} \prod_{s'=1}^{r} (X_0 - \alpha_{s'} X_1)^{m_{s'}}, \end{align*} a factorisation of $f$ into linear forms in $k[X_0, X_1]$. The total degree is \begin{align*} m_\infty + \sum_{s'=1}^{r} m_{s'} = m_\infty + r' = c, \end{align*} as required. \textbf{Interpretation.} The linear factor $(X_0 - \alpha_{s'} X_1)$ vanishes at the point $[\alpha_{s'} : 1] \in \mathbb{P}^1_k$, and the factor $X_1$ (present when $m_\infty \geq 1$) vanishes at the point $[1:0] \in \mathbb{P}^1_k$. Since the $\alpha_{s'}$ are distinct and all non-zero, the points $[\alpha_1 : 1], \ldots, [\alpha_r : 1], [1:0]$ are pairwise distinct in $\mathbb{P}^1_k$. The zero set $V_{\mathbb{P}^1_k}(f)$ therefore consists of exactly $r$ distinct points if $m_\infty = 0$, or $r + 1$ distinct points if $m_\infty \geq 1$, and the sum of multiplicities of the linear factors is exactly $c$.[/step]

[guided]Binary forms over an algebraically closed field factor uniquely into linear forms — this is the homogeneous version of the *Fundamental Theorem of Algebra*. Why does it hold? Dehomogenise in a chart where the polynomial does not identically vanish — here the chart $\{X_1 \neq 0\}$, chosen because $f(0,1) = a_{0,c,0} \neq 0$ (arranged in Step 1). The resulting univariate polynomial $\hat f(s) = f(s, 1)$ has some degree $r' \leq c$, not necessarily equal to $c$: the degree drop $m_\infty := c - r'$ equals the multiplicity of the "point at infinity" $[1:0]$ of the chosen chart, which appears as a factor of $X_1$ in the rehomogenisation. \textbf{Which coefficient is the leading coefficient?} This is the subtlety that a careless write-up easily misses. The \emph{leading coefficient} of $\hat f$ in the variable $s$ is the coefficient of $s^{r'}$, which is $a_{r',\, c - r',\, 0}$. In contrast, the constant $a_{0,c,0}$ is the \emph{constant term} of $\hat f$ (the coefficient of $s^0$), which plays the dual role of being $f(0,1) = F(0,1,0)$ — the evaluation of $F$ at the point $[0:1:0]$ that we arranged to avoid $C$. When $r' = c$ (i.e. $a_{c,0,0} \neq 0$), the leading coefficient is $a_{c,0,0}$ and $m_\infty = 0$ (no factor of $X_1$ in the factorisation); when $r' < c$, the leading coefficient is some $a_{r',\, c-r',\, 0}$ with $r' < c$ and $m_\infty \geq 1$. \textbf{Why choose the chart $\{X_1 \neq 0\}$ rather than $\{X_0 \neq 0\}$?} The chart is determined by where the form does not vanish: $\hat f(s) = f(s,1)$ has a nonzero value at $s = 0$ precisely because $f(0,1) = a_{0,c,0} \neq 0$. Using $\{X_0 \neq 0\}$ instead would require $a_{c,0,0} = F(1,0,0) \neq 0$, which we did not arrange — it would require a different coordinate-change step. Step 1 was designed around the chart $\{X_1 \neq 0\}$. \textbf{Rehomogenising.} Each linear factor $(s - \alpha_{s'})$ of $\hat f$ becomes $(X_0 - \alpha_{s'} X_1)/X_1$ when substituting $s = X_0/X_1$, and multiplying through by $X_1^c$ distributes as follows: $X_1^{r'}$ cancels the denominators of the $r'$ linear factors, leaving $X_1^{c - r'} = X_1^{m_\infty}$ as the "at-infinity" multiplier. This is exactly the rule: a missing multiplicity in the dehomogenisation becomes a factor of $X_1$ in the rehomogenisation. The factorisation of $f$ into $c$ linear forms (counted with multiplicity) is the homogeneous statement that a binary form of degree $c$ has exactly $c$ zeros counted with multiplicity in $\mathbb{P}^1_k$. The number of distinct zeros is $r$ or $r+1$, always at most $c$, with equality iff all multiplicities are $1$.[/guided]

[step:Identify the intersection multiplicity $m_p(L, C)$ with the multiplicity of the corresponding root] For each $s \in \{1, \ldots, r\}$, the linear factor $(X_0 - \alpha_s X_1)^{m_s}$ contributes a zero of $f$ at the point $[\alpha_s : 1] \in \mathbb{P}^1_k$, which corresponds via $\iota$ to the point $p_s := [\alpha_s : 1 : 0] \in L \cap C$. We claim \begin{align*} m_{p_s}(L, C) = m_s. \end{align*} This is the standard local computation. Working in the affine chart $U_1 = \{X_1 \neq 0\}$ with affine coordinates $(u, v) = (X_0/X_1, X_2/X_1)$, the line $L = V(X_2)$ becomes $V(v)$, and the curve $C = V(F)$ becomes $V(\tilde F)$ where $\tilde F(u, v) := F(u, 1, v)$. The local ring $\mathcal{O}_{\mathbb{A}^2_k, p_s}$ has maximal ideal generated by $(u - \alpha_s, v)$, and the intersection multiplicity is \begin{align*} m_{p_s}(L, C) = \dim_k \mathcal{O}_{\mathbb{A}^2_k, p_s} / (v, \tilde F(u, v)) = \dim_k \mathcal{O}_{\mathbb{A}^1_k, \alpha_s} / (\tilde F(u, 0)), \end{align*} using the substitution $v = 0$ to descend to $\mathbb{A}^1_k$. Now \begin{align*} \tilde F(u, 0) = F(u, 1, 0) = f(u, 1) = \hat f(u) = a_{r',\, c - r',\, 0} \prod_{s'=1}^{r} (u - \alpha_{s'})^{m_{s'}}, \end{align*} writing the factorisation of Step 4 in the affine variable $u = X_0/X_1$ (no factor of $X_1$ appears because we have set $X_1 = 1$). The localisation at the point $u = \alpha_s$ extracts only the $(u - \alpha_s)^{m_s}$ factor: the leading constant $a_{r',\, c - r',\, 0} \in k^\times$ is a unit in the local ring, and the other factors $(u - \alpha_{s'})$ with $s' \neq s$ evaluate to $\alpha_s - \alpha_{s'} \neq 0$ (since the $\alpha$'s are distinct), hence become units in $\mathcal{O}_{\mathbb{A}^1_k, \alpha_s}$. Therefore \begin{align*} \mathcal{O}_{\mathbb{A}^1_k, \alpha_s} / (\hat f(u)) = \mathcal{O}_{\mathbb{A}^1_k, \alpha_s} / ((u - \alpha_s)^{m_s}) \end{align*} is a $k$-vector space of dimension $m_s$, with basis $\{1, (u - \alpha_s), \ldots, (u - \alpha_s)^{m_s - 1}\}$. Therefore $m_{p_s}(L, C) = m_s$. A symmetric argument applies to the point $p_\infty := [1:0:0] \in L$ (if $m_\infty \geq 1$, i.e.\ if this point is in $L \cap C$): in the chart $U_0 = \{X_0 \neq 0\}$, the relevant local computation yields $m_{p_\infty}(L, C) = m_\infty$. [/step]

[step:Sum the multiplicities to obtain the Bézout count] Combining Steps 3, 4, and 5: the intersection $L \cap C$ consists of the points \begin{align*} \{p_s = [\alpha_s : 1 : 0] : s = 1, \ldots, r\} \cup \{p_\infty = [1:0:0] \text{ if } m_\infty \geq 1\}, \end{align*} each appearing with intersection multiplicity equal to its root multiplicity in the factorisation of $f$. Therefore \begin{align*} \sum_{p \in L \cap C} m_p(L, C) = \sum_{s=1}^{r} m_s + m_\infty = c, \end{align*} the total degree of the factorisation of $f$. For the cardinality bound, every multiplicity is at least $1$, so \begin{align*} |L \cap C| = r + \mathbb{1}_{m_\infty \geq 1} \leq \sum_{s=1}^{r} m_s + m_\infty = c. \end{align*} This completes the proof. [/step]