[proofplan] The crux is that the ratio $L/L'$ is a globally defined nonzero rational function on $C$ (well-defined because neither $L$ nor $L'$ vanishes identically on $C$, so $L'$ is not the zero element of $K(C)$ and the ratio makes sense in the function field), and its principal divisor is exactly $\operatorname{div}(L) - \operatorname{div}(L')$. By [Principal Divisors Have Degree Zero](/theorems/2177), the degree of any principal divisor is $0$, giving $\deg \operatorname{div}(L) - \deg \operatorname{div}(L') = 0$. [/proofplan] [step:Define the rational function $L/L'$ on $C$] The linear forms $L, L' \in k[X_0, \ldots, X_n]_1$ are homogeneous of degree $1$, so the ratio \begin{align*} \frac{L}{L'} \in k(X_0, \ldots, X_n) \end{align*} is homogeneous of degree $0$ as a rational function on $\mathbb{P}^n_k$. Restriction to $C$ gives a rational function \begin{align*} \frac{L}{L'}\bigg|_C \in K(C), \end{align*} provided the restriction is nonzero in $K(C)$. We verify nonvanishing: the restriction of $L'$ to $C$ is a nonzero element of the homogeneous coordinate ring of $C$, because $C \not\subset V(L')$ means there exists $p \in C$ with $L'(p) \neq 0$; in particular, $L'|_C$ is not the zero rational function on $C$. Similarly $L|_C \neq 0$ in $K(C)$. Hence the ratio $L/L'|_C$ is a nonzero element of $K(C)$: \begin{align*} f := \frac{L}{L'}\bigg|_C \in K(C)^\times = \mathcal{O}_C(\eta)^\times. \end{align*} [/step] [step:Compute the local order of $f = L/L'$ at every closed point of $C$] Fix a closed point $p \in C$. We compute $\operatorname{ord}_p(f)$ in terms of the local orders of $L|_C$ and $L'|_C$ at $p$. Choose a homogeneous linear form $L_0$ on $\mathbb{P}^n_k$ with $L_0(p) \neq 0$ (such a form exists because $p$ is a single point and we can choose any standard coordinate $X_i$ with $X_i(p) \neq 0$, possible because $p \in \mathbb{P}^n_k$ has at least one nonzero homogeneous coordinate). Then $L/L_0$ and $L'/L_0$ are degree-$0$ rational functions on $\mathbb{P}^n_k$ that are regular at $p$ (because $L_0(p) \neq 0$). Restricting to $C$: \begin{align*} \frac{L}{L_0}\bigg|_C, \quad \frac{L'}{L_0}\bigg|_C \in \mathcal{O}_{C, p}. \end{align*} The order of vanishing of $L|_C$ at $p$, by definition of the divisor of a homogeneous form on a projective curve, is \begin{align*} \operatorname{ord}_p(L|_C) := \operatorname{ord}_p\!\left(\frac{L}{L_0}\bigg|_C\right) \in \mathbb{Z}_{\geq 0}, \end{align*} and is independent of the choice of $L_0$ (a different choice $L_0'$ with $L_0'(p) \neq 0$ differs by the unit $L_0/L_0'$, which has order $0$ at $p$). Similarly for $L'|_C$. Now in $K(C)$: \begin{align*} f = \frac{L}{L'} = \frac{L/L_0}{L'/L_0}, \end{align*} so by additivity of $\operatorname{ord}_p$ on $K(C)^\times$: \begin{align*} \operatorname{ord}_p(f) = \operatorname{ord}_p\!\left(\frac{L}{L_0}\bigg|_C\right) - \operatorname{ord}_p\!\left(\frac{L'}{L_0}\bigg|_C\right) = \operatorname{ord}_p(L|_C) - \operatorname{ord}_p(L'|_C). \end{align*} [/step] [step:Translate the local identity into divisor language] Summing the local identity over all $p \in C$ and weighting by the formal symbol $[p]$: \begin{align*} \operatorname{div}(f) = \sum_{p \in C} \operatorname{ord}_p(f) \cdot [p] = \sum_{p \in C} \operatorname{ord}_p(L|_C) \cdot [p] - \sum_{p \in C} \operatorname{ord}_p(L'|_C) \cdot [p] = \operatorname{div}(L) - \operatorname{div}(L'), \end{align*} where the equality is in the divisor group $\operatorname{Div}(C)$. (Each of the three sums is a finite formal sum: $\operatorname{div}(f)$ has finite support by part (2) of [Sum of Orders Is Zero](/theorems/2176), and $\operatorname{div}(L), \operatorname{div}(L')$ have finite support because $L|_C, L'|_C$ are nonzero homogeneous forms on the projective curve $C$, so their zero loci are zero-dimensional, hence finite.) [/step] [step:Apply the degree-zero theorem for principal divisors to conclude] Since $f \in \mathcal{O}_C(\eta)^\times$, the divisor $\operatorname{div}(f)$ is principal. By [Principal Divisors Have Degree Zero](/theorems/2177) (whose hypotheses — $C$ a smooth projective irreducible curve over an algebraically closed field, $f$ a nonzero rational function — are met): \begin{align*} \deg \operatorname{div}(f) = 0. \end{align*} Combining with Step 3: \begin{align*} 0 = \deg \operatorname{div}(f) = \deg(\operatorname{div}(L) - \operatorname{div}(L')) = \deg \operatorname{div}(L) - \deg \operatorname{div}(L'), \end{align*} where the last equality uses additivity of the degree homomorphism $\deg : \operatorname{Div}(C) \to \mathbb{Z}$. Rearranging: \begin{align*} \deg \operatorname{div}(L) = \deg \operatorname{div}(L'). \end{align*} [guided] The proof is a paradigmatic application of [Principal Divisors Have Degree Zero](/theorems/2177): two divisors that differ by a principal divisor have the same degree. Let us understand the moving parts. \textbf{Why is $L/L'$ a rational function?} A linear form $L$ on $\mathbb{P}^n_k$ does not define a global function on $\mathbb{P}^n_k$ — it is homogeneous of degree $1$, and a function would need to be homogeneous of degree $0$. But the ratio $L/L'$ of two linear forms is homogeneous of degree $0$, hence a rational function. The condition $C \not\subset V(L')$ ensures that $L'$ does not vanish identically on $C$, so the ratio is a well-defined element of $K(C)^\times$ and not $0/0$ everywhere. Without this condition, the construction would fail. \textbf{Why is $\operatorname{div}(L/L') = \operatorname{div}(L) - \operatorname{div}(L')$?} The order $\operatorname{ord}_p$ on $K(C)^\times$ is a homomorphism $K(C)^\times \to \mathbb{Z}$, so $\operatorname{ord}_p(L/L') = \operatorname{ord}_p(L) - \operatorname{ord}_p(L')$. The local-to-global passage to divisors is just summing this point-by-point identity over all $p$. But wait — $L$ and $L'$ are not themselves elements of $K(C)$; they are homogeneous forms. To make sense of $\operatorname{ord}_p(L|_C)$, we de-homogenise: choose a third linear form $L_0$ with $L_0(p) \neq 0$, and define $\operatorname{ord}_p(L|_C) := \operatorname{ord}_p(L/L_0|_C)$ as the order of the dehomogenised function $L/L_0$ on $C$ at $p$. This is independent of the choice of $L_0$ because two such choices differ by a unit at $p$. With this convention, the homomorphism property $\operatorname{ord}_p(LL'^{-1}) = \operatorname{ord}_p(L) - \operatorname{ord}_p(L')$ holds, and we obtain $\operatorname{div}(L/L') = \operatorname{div}(L) - \operatorname{div}(L')$. \textbf{The degree comparison.} Now apply $\deg$ to both sides. The function $f = L/L'$ is in $K(C)^\times$, so $\operatorname{div}(f)$ is principal, so $\deg \operatorname{div}(f) = 0$ by Theorem 2177. The right-hand side gives $\deg \operatorname{div}(L) - \deg \operatorname{div}(L')$, and we conclude these are equal. \textbf{Strategic significance.} This proves that the integer $\deg \operatorname{div}(L)$ — the number of intersection points of $C$ with a hyperplane $V(L)$, counted with multiplicities — is independent of the chosen hyperplane (provided the hyperplane does not contain $C$). This common value is the \emph{degree} of the curve $C$ in $\mathbb{P}^n_k$. Without [Principal Divisors Have Degree Zero](/theorems/2177), this fundamental invariant would not be well-defined. [/guided] [/step]