[proofplan] The quotient $F/G$ has homogeneous degree zero, so it is invariant under rescaling homogeneous coordinates and therefore gives a rational function on the irreducible variety $X$. To compute its principal divisor, we work at the generic point of each prime divisor $Y\subset X$ and trivialize $\mathcal O_X(d)$ on a standard affine chart meeting that generic point. In that local trivialization the sections $F$ and $G$ become ordinary regular functions, so the valuation identity $v_Y(f/g)=v_Y(f)-v_Y(g)$ gives the coefficient of $Y$ in the divisor of $F/G$. Summing over all prime divisors gives the displayed divisor formula, and linear equivalence follows because the difference is principal. [/proofplan]

[step:Construct the degree-zero quotient on affine charts]For each $i\in\{0,\dots,n\}$, let \begin{align*} D_+(X_i):=\{p\in \mathbb P_k^n:X_i(p)\neq 0\} \end{align*} denote the $i$-th standard affine open subset of projective space. By the [standard affine cover of projective space](/theorems/9430) [citetheorem:9430], these opens cover $\mathbb P_k^n$. On $X\cap D_+(X_i)$, define the local representatives of the two sections by \begin{align*} f_i:=F/X_i^d \end{align*} and \begin{align*} g_i:=G/X_i^d. \end{align*} These are regular functions on $X\cap D_+(X_i)$. On the nonempty open subset where $g_i$ is nonzero, the quotient $f_i/g_i$ is a regular function, and in the common function field $k(X)$ it is independent of $i$ because \begin{align*} \frac{f_i}{g_i}=\frac{F/X_i^d}{G/X_i^d}=\frac{F}{G}. \end{align*} Thus the compatible chartwise quotients define an element of $k(X)$. Since the restriction of $F$ to $X$ is nonzero and $X$ is integral, this element is nonzero in $k(X)$. We denote this rational function by $F/G$.[/step]

[guided]The quotient $F/G$ should be constructed as a rational function, not as a globally defined regular function. For each index $i\in\{0,\dots,n\}$, define \begin{align*} D_+(X_i):=\{p\in \mathbb P_k^n:X_i(p)\neq 0\}. \end{align*} The standard [affine cover of projective space](/theorems/2129) [citetheorem:9430] says that the opens $D_+(X_i)$ cover $\mathbb P_k^n$, so their intersections with $X$ cover $X$. On $X\cap D_+(X_i)$, the line bundle $\mathcal O_X(d)$ is trivialized by $X_i^d$. In that trivialization, the section defined by $F$ is represented by the regular function \begin{align*} f_i:=F/X_i^d, \end{align*} and the section defined by $G$ is represented by the regular function \begin{align*} g_i:=G/X_i^d. \end{align*} Because the restriction of $G$ to $X$ is not the zero section and $X$ is integral, $g_i$ is not zero in the local function field wherever this chart meets the generic point. Hence $f_i/g_i$ defines a rational function on $X\cap D_+(X_i)$. The chartwise definitions agree in the function field. Indeed, for every chart on which the expression is written, cancellation of the same trivializing factor gives \begin{align*} \frac{f_i}{g_i}=\frac{F/X_i^d}{G/X_i^d}=\frac{F}{G}. \end{align*} Thus the local quotients glue at the level of rational functions and define a single element of $k(X)$. Since the restriction of $F$ to $X$ is also not the zero section, this element is not the zero rational function. We denote it by $F/G$.[/guided]

[step:Compute the coefficient along each prime divisor] Let $Y\subset X$ be a prime divisor, and let $\eta_Y\in X$ denote its generic point. For a homogeneous form $H\in S$, write \begin{align*} D_+(H):=\{p\in \mathbb P_k^n:H(p)\neq 0\}. \end{align*} By the standard affine cover of projective space [citetheorem:9430], the standard affine opens $D_+(X_i)$ cover $\mathbb P_k^n$. Choose an index $i\in\{0,\dots,n\}$ such that \begin{align*} \eta_Y\in X\cap D_+(X_i). \end{align*} On the affine chart $D_+(X_i)$, the section of $\mathcal O_X(d)$ defined by $F$ is represented by the regular function \begin{align*} f_i:=\frac{F}{X_i^d}, \end{align*} and the section defined by $G$ is represented by the regular function \begin{align*} g_i:=\frac{G}{X_i^d}. \end{align*} These functions are elements of the local ring $\mathcal O_{X,\eta_Y}$. Because $X$ is nonsingular, the local ring $\mathcal O_{X,\eta_Y}$ at the generic point of the prime divisor $Y$ is a one-dimensional regular local ring, hence a discrete valuation ring. Let \begin{align*} v_Y:k(X)^\times\to \mathbb Z \end{align*} be its associated valuation. For a Weil divisor $D=\sum_Z m_Z Z$ on $X$, where the sum ranges over prime divisors $Z\subset X$, define $\operatorname{coeff}_Y(D):=m_Y$. By definition of the zero divisor of a section, \begin{align*} \operatorname{coeff}_Y((F)_0|_X)=v_Y(f_i) \end{align*} and \begin{align*} \operatorname{coeff}_Y((G)_0|_X)=v_Y(g_i). \end{align*} Moreover, in $k(X)$ one has \begin{align*} \frac{f_i}{g_i}=\frac{F/X_i^d}{G/X_i^d}=\frac{F}{G}. \end{align*} Using the valuation rule for a quotient in a discrete valuation field, we obtain \begin{align*} v_Y(F/G)=v_Y(f_i/g_i)=v_Y(f_i)-v_Y(g_i). \end{align*} Therefore \begin{align*} \operatorname{coeff}_Y(\operatorname{div}(F/G))=\operatorname{coeff}_Y((F)_0|_X)-\operatorname{coeff}_Y((G)_0|_X). \end{align*} [/step]

[step:Sum the local coefficient identities to obtain the divisor formula] The preceding coefficient identity holds for every prime divisor $Y\subset X$. The principal divisor of $F/G$ is, by definition, \begin{align*} \operatorname{div}(F/G)=\sum_Y v_Y(F/G)Y, \end{align*} where the sum ranges over prime divisors $Y\subset X$ and has finite support. Likewise, \begin{align*} (F)_0|_X=\sum_Y \operatorname{coeff}_Y((F)_0|_X)Y \end{align*} and \begin{align*} (G)_0|_X=\sum_Y \operatorname{coeff}_Y((G)_0|_X)Y. \end{align*} Since the coefficients agree prime divisor by prime divisor, we get \begin{align*} \operatorname{div}(F/G)=(F)_0|_X-(G)_0|_X. \end{align*} [/step]

[step:Conclude linear equivalence from the principal divisor] By definition, two divisors on $X$ are linearly equivalent if their difference is the divisor of a nonzero rational function on $X$. The formula already proved gives \begin{align*} (F)_0|_X-(G)_0|_X=\operatorname{div}(F/G). \end{align*} Since $F/G\in k(X)^\times$, the right-hand side is a principal divisor. Hence \begin{align*} (F)_0|_X\sim (G)_0|_X. \end{align*} This proves both the precise divisor identity and the asserted linear equivalence. [/step]