[proofplan] The strategy is to bound the cardinality of $C \cap C'$ by the degree of an effective divisor on $C$ that has support containing $C \cap C'$. Write $C' = V(G)$ for a homogeneous polynomial $G \in k[X_0, X_1, X_2]$ of degree $\deg(C')$. The restriction of $G$ to $C$ defines an effective divisor $\operatorname{div}(G)$ on $C$ — the divisor of zeros of $G|_C$ — whose support contains $C \cap C'$. Since each zero contributes at least $1$ to the order, $|C \cap C'| \leq \deg \operatorname{div}(G)$. By [Degree of Hyperplane Sections](/theorems/2178), the degree of the divisor of any nonzero homogeneous form of degree $d$ on $C$ is independent of the form (provided $C$ is not contained in its zero locus) and equals $d \cdot \deg(C)$ — applied with $d = 1$ this is the definition of $\deg(C)$, and the case $d = \deg(C')$ follows by writing $G$ as a product of $\deg(C')$ linear factors over an algebraic closure (or more directly by the multiplicativity $\deg \operatorname{div}(L_1 \cdots L_d) = \sum \deg \operatorname{div}(L_i)$). [/proofplan] [step:Set up the homogeneous polynomial $G$ defining $C'$] Write $C' = V(G)$ for some nonzero homogeneous polynomial $G \in k[X_0, X_1, X_2]_{\deg(C')}$ of degree $\deg(C')$. Since $C \neq C'$ and $C$ is irreducible, $C \not\subset C' = V(G)$ (otherwise $C \subset C'$ would force $C \subset C'$ as closed subvarieties; since both are curves and $C$ is irreducible, irreducibility plus inclusion plus dimension equality would imply $C \subset C'$ is a component of $C'$, but if $C' \neq C$ then either $C'$ contains additional components — possible but the inclusion $C \subset C'$ already gives $C = C'$ in the irreducible case if dimensions match, contradiction). Concretely, the restriction $G|_C$ is a nonzero element of the homogeneous coordinate ring of $C$. [/step] [step:Define the divisor $\operatorname{div}(G)$ on $C$] For each closed point $p \in C$, define $\operatorname{ord}_p(G|_C)$ as follows. Choose a homogeneous linear form $L_0 \in k[X_0, X_1, X_2]_1$ with $L_0(p) \neq 0$ (such $L_0$ exists because $p$ has at least one nonzero homogeneous coordinate, so a corresponding $X_i$ works). Then $G/L_0^{\deg(C')}$ is a degree-$0$ rational function on $\mathbb{P}^2_k$, regular at $p$, and its restriction to $C$ lies in $\mathcal{O}_{C, p}$. Define \begin{align*} \operatorname{ord}_p(G|_C) := \operatorname{ord}_p\!\left(\frac{G}{L_0^{\deg(C')}}\bigg|_C\right) \in \mathbb{Z}_{\geq 0}, \end{align*} which is independent of the choice of $L_0$ (any other choice differs by a unit in $\mathcal{O}_{C, p}$). The integer $\operatorname{ord}_p(G|_C)$ is non-negative because $G/L_0^{\deg(C')}|_C \in \mathcal{O}_{C, p}$ is regular at $p$. The associated divisor on $C$ is \begin{align*} \operatorname{div}(G) := \sum_{p \in C} \operatorname{ord}_p(G|_C) \cdot [p] \in \operatorname{Div}(C). \end{align*} This is a finite formal sum because the rational function $G/L_0^{\deg(C')}|_C$ on $C$ has finitely many zeros (its order is nonzero on a finite set, by part (2) of [Sum of Orders Is Zero](/theorems/2176) applied to the nonzero rational function $G/L_0^{\deg(C')}|_C$; only the zero contributions survive in the formula above since the function is regular). The divisor $\operatorname{div}(G)$ is effective: all coefficients are non-negative. [/step] [step:Bound $|C \cap C'|$ by the degree of $\operatorname{div}(G)$] The intersection $C \cap C' = \{p \in C : G(p) = 0\}$ is exactly the support of $\operatorname{div}(G)$: \begin{align*} \operatorname{Supp}(\operatorname{div}(G)) = \{p \in C : \operatorname{ord}_p(G|_C) > 0\} = \{p \in C : G(p) = 0\} = C \cap C'. \end{align*} The first equality is the definition of the support of an effective divisor, the second is because $\operatorname{ord}_p(G|_C) > 0$ if and only if $(G/L_0^{\deg(C')})|_C$ vanishes at $p$ if and only if $G(p) = 0$ (since $L_0(p) \neq 0$). Each $p \in C \cap C'$ contributes a coefficient $\operatorname{ord}_p(G|_C) \geq 1$ in $\operatorname{div}(G)$. Hence \begin{align*} \deg \operatorname{div}(G) = \sum_{p \in C} \operatorname{ord}_p(G|_C) = \sum_{p \in C \cap C'} \operatorname{ord}_p(G|_C) \geq \sum_{p \in C \cap C'} 1 = |C \cap C'|. \end{align*} The intersection $C \cap C'$ is finite because $C \neq C'$ implies $C \cap C'$ is a proper closed subset of the irreducible curve $C$, hence zero-dimensional. [/step] [step:Compute $\deg \operatorname{div}(G) = \deg(C) \cdot \deg(C')$] Let $d := \deg(C')$, so $G \in k[X_0, X_1, X_2]_d$ is homogeneous of degree $d$. Over the algebraically closed field $k$, the homogeneous polynomial $G$ in three variables does not in general factor into linear forms (for $d \geq 2$ and three variables, it factors into irreducible factors of various degrees), so we cannot directly reduce to linear forms. Instead, we proceed by comparing $G$ to a product of $d$ linear forms in general position. Choose $d$ distinct hyperplanes $V(L_1), \ldots, V(L_d) \subset \mathbb{P}^2_k$, each not containing $C$ (such a choice exists: the set of linear forms whose vanishing locus contains $C$ is a proper subspace of $k[X_0, X_1, X_2]_1$, since $C$ is two-dimensional in projective ambient terms... we elaborate). Set $H := L_1 L_2 \cdots L_d \in k[X_0, X_1, X_2]_d$, also a homogeneous polynomial of degree $d$, and not vanishing identically on $C$ (since none of the $L_i$ does, and $C$ is irreducible, the product does not vanish on $C$). Both $G/L_0^d$ and $H/L_0^d$ are degree-$0$ rational functions on $\mathbb{P}^2_k$, regular wherever $L_0 \neq 0$. Their ratio $G/H$ is a degree-$0$ rational function on $\mathbb{P}^2_k$. Its restriction to $C$: \begin{align*} \frac{G}{H}\bigg|_C \in K(C)^\times, \end{align*} is nonzero (both $G|_C$ and $H|_C$ are nonzero rational functions on $C$). The principal divisor identity from the previous theorem applies: \begin{align*} \operatorname{div}\!\left(\frac{G}{H}\bigg|_C\right) = \operatorname{div}(G) - \operatorname{div}(H), \end{align*} by additivity of $\operatorname{ord}_p$ on $K(C)^\times$ and the same dehomogenisation argument used in the proof of [Degree of Hyperplane Sections](/theorems/2178). By [Principal Divisors Have Degree Zero](/theorems/2177), the left-hand side has degree $0$. Hence \begin{align*} \deg \operatorname{div}(G) = \deg \operatorname{div}(H). \end{align*} For the product $H = L_1 \cdots L_d$, we have additivity: \begin{align*} \operatorname{ord}_p(H|_C) = \sum_{i=1}^{d} \operatorname{ord}_p(L_i|_C) \quad \text{for each } p \in C, \end{align*} because $H/L_0^d|_C = \prod_i (L_i/L_0)|_C$ and $\operatorname{ord}_p$ is additive on $K(C)^\times$. Summing over $p$: \begin{align*} \operatorname{div}(H) = \sum_{i=1}^{d} \operatorname{div}(L_i), \end{align*} and taking degrees: \begin{align*} \deg \operatorname{div}(H) = \sum_{i=1}^{d} \deg \operatorname{div}(L_i). \end{align*} By [Degree of Hyperplane Sections](/theorems/2178), each $\deg \operatorname{div}(L_i)$ equals the common value $\deg(C)$ (this common value is the definition of the degree of $C$ as a projective curve). Hence \begin{align*} \deg \operatorname{div}(H) = d \cdot \deg(C) = \deg(C') \cdot \deg(C), \end{align*} and therefore \begin{align*} \deg \operatorname{div}(G) = \deg(C) \cdot \deg(C'). \end{align*} [guided] This step is the bridge from the algebraic quantity $\deg \operatorname{div}(G)$ to the geometric quantity $\deg(C) \cdot \deg(C')$. The argument is a two-step deformation through a comparison polynomial $H$. \textbf{Step (a): replace $G$ by $H$ via a principal divisor.} The polynomials $G$ and $H = L_1 \cdots L_d$ are both homogeneous of the same degree $d = \deg(C')$, and neither vanishes identically on $C$. The ratio $G/H$ is then a nonzero rational function on $C$ — degree-$0$ as a rational function on the ambient $\mathbb{P}^2_k$. Its principal divisor on $C$ is $\operatorname{div}(G/H|_C) = \operatorname{div}(G) - \operatorname{div}(H)$, which has degree $0$ by [Principal Divisors Have Degree Zero](/theorems/2177). Therefore $\deg \operatorname{div}(G) = \deg \operatorname{div}(H)$. This is the crucial use of the theorem from the previous section: the degree of the divisor cut out by a homogeneous form of degree $d$ on $C$ depends only on $d$, not on the specific form. \textbf{Step (b): compute $\deg \operatorname{div}(H)$ for $H$ a product of linear forms.} Writing $H = L_1 \cdots L_d$ as a product of $d$ linear forms, the order at each $p \in C$ decomposes additively: $\operatorname{ord}_p(H|_C) = \sum_i \operatorname{ord}_p(L_i|_C)$. Summing over $p$, the divisor decomposes: $\operatorname{div}(H) = \sum_i \operatorname{div}(L_i)$. Each summand $\operatorname{div}(L_i)$ is a hyperplane-section divisor and has degree equal to $\deg(C)$ by [Degree of Hyperplane Sections](/theorems/2178) — this common value is the very definition of $\deg(C)$. Therefore $\deg \operatorname{div}(H) = d \cdot \deg(C) = \deg(C') \cdot \deg(C)$. \textbf{Why this defines $\deg(C)$.} The number $\deg \operatorname{div}(L)$ for any linear form $L$ not vanishing on $C$ is a well-defined invariant of $C$ (Theorem 2178). This invariant is, by definition, $\deg(C)$. Geometrically, it counts the number of intersection points of $C$ with a hyperplane (counted with multiplicities) in $\mathbb{P}^2_k$. \textbf{Why the comparison polynomial $H$ exists.} Choosing $d$ distinct linear forms $L_i$ none of which vanishes on $C$ is possible because the linear forms vanishing on $C$ form a proper linear subspace of $k[X_0, X_1, X_2]_1 \cong k^3$ (this subspace is $\{0\}$ when $C \not\subset V(L)$ for any nonzero $L$, which holds whenever $C$ is not contained in any hyperplane; for $\mathbb{P}^2_k$, every irreducible curve $C$ is by hypothesis not the empty set in any affine chart and is two-dimensional in homogeneous coordinates, so it cannot be contained in a hyperplane unless it equals one — in which case the formula $|C \cap C'| \leq \deg(C) \deg(C')$ is trivially handled). Concretely: pick any linear form $L$ with $L|_C \neq 0$; then the family $\{L + \lambda X_0\}_{\lambda \in k}$ provides infinitely many such linear forms, and we choose $d$ of them. [/guided] [/step] [step:Combine to obtain Bézout's bound] Combining Steps 3 and 4: \begin{align*} |C \cap C'| \leq \deg \operatorname{div}(G) = \deg(C) \cdot \deg(C'). \end{align*} This is the desired inequality. [guided] The bound is an inequality, not an equality, because the divisor $\operatorname{div}(G)$ counts each intersection point $p$ with its multiplicity $\operatorname{ord}_p(G|_C)$, while $|C \cap C'|$ counts each point only once. The strict inequality $|C \cap C'| < \deg(C) \cdot \deg(C')$ occurs precisely when some intersection point has multiplicity $\geq 2$ — i.e., $C$ and $C'$ are tangent to a higher order at $p$. The full Bézout theorem (with multiplicities) states equality: \begin{align*} \sum_{p \in C \cap C'} (C \cdot C')_p = \deg(C) \cdot \deg(C'), \end{align*} where $(C \cdot C')_p$ is the local intersection multiplicity. For smooth $C$, this multiplicity coincides with $\operatorname{ord}_p(G|_C)$, so $\deg \operatorname{div}(G) = \sum_p (C \cdot C')_p$, and the inequality of this proof becomes equality when intersection points are weighted with multiplicities. The version we have proved here is the unweighted form: the number of distinct points in $C \cap C'$ is at most $\deg(C) \cdot \deg(C')$. [/guided] [/step]