[proofplan] Choose a nonzero homogeneous linear form $L\in k[x_0,\dots,x_n]_1$ defining $H$. Its restriction to $X$ is a nonzero global section of $\mathcal O_X(1)$ because $X$ is not contained in $H$. On local trivializations of $\mathcal O_X(1)$ this section is represented by regular functions, and changing trivializations only multiplies those functions by units. Therefore the orders of vanishing along prime divisors of $X$ are well-defined and nonnegative, which is exactly the effectivity of the hyperplane section divisor. [/proofplan]

[step:Restrict the defining linear form to a nonzero section on $X$] Let $S:=k[x_0,\dots,x_n]$ be the homogeneous coordinate ring of $\mathbb P_k^n$. By the theorem statement, the hyperplane has the form $H=V_+(L)\subset \mathbb P_k^n$ for a nonzero homogeneous linear form $L\in S_1$. Let $H^0(X,\mathcal O_X(1))$ denote the $k$-[vector space](/page/Vector%20Space) of global sections of the invertible sheaf $\mathcal O_X(1)$, and let $s_L\in H^0(X,\mathcal O_X(1))$ denote the restriction to $X$ of the global section of $\mathcal O_{\mathbb P_k^n}(1)$ represented by $L$. We claim that $s_L$ is not the zero section. If $s_L=0$, then $L$ vanishes at every point of $X$, so $X\subset V_+(L)=H$, contradicting the hypothesis $X\not\subset H$. Hence $s_L\ne 0$. [/step]

[step:Represent the restricted section by regular local equations]Let $U\subset X$ be an open subset on which $\mathcal O_X(1)$ is trivial, and choose a local frame $e_U:\mathcal O_U\longrightarrow \mathcal O_X(1)|_U$ for $\mathcal O_X(1)$ over $U$. Let $\mathbb A_k^1:=\operatorname{Spec} k[t]$ denote the affine line over $k$. Since $e_U$ is an isomorphism of invertible sheaves on $U$, there is a unique regular function $f_U:U\longrightarrow \mathbb A_k^1$ such that $s_L|_U=e_U(f_U)$. Thus, locally on $X$, the hyperplane section is cut out by the regular function $f_U$.[/step]

[guided]The section $s_L$ is a section of a line bundle, not globally a regular function on $X$. To measure its vanishing, we first make it into a regular function locally. Since $\mathcal O_X(1)$ is an invertible sheaf, it is locally free of rank one; hence for every point of $X$ there is an open neighbourhood on which it is trivial. Choose such an [open set](/page/Open%20Set) $U\subset X$ and choose a local frame \begin{align*} e_U:\mathcal O_U\longrightarrow \mathcal O_X(1)|_U. \end{align*} This means that every local section of $\mathcal O_X(1)$ over $U$ can be written uniquely as the frame $e_U$ multiplied by a regular function on $U$. Applying this to $s_L|_U$, there is a unique regular function \begin{align*} f_U:U\longrightarrow \mathbb A_k^1 \end{align*} with \begin{align*} s_L|_U=e_U(f_U). \end{align*} This function $f_U$ is the local equation of the hyperplane section on $U$. The point of using a local frame is that orders of vanishing are defined for elements of local rings, and $f_U$ gives precisely such an element at each point of $U$.[/guided]

[step:Check that local equations have frame-independent orders] Let $V\subset X$ be another open subset on which $\mathcal O_X(1)$ is trivial, with local frame \begin{align*} e_V:\mathcal O_V\longrightarrow \mathcal O_X(1)|_V, \end{align*} and let $f_V:V\longrightarrow \mathbb A_k^1$ be the regular function satisfying $s_L|_V=e_V(f_V)$. On $U\cap V$, the two frames differ by a unit. Thus there is a regular unit $u_{UV}\in \mathcal O_X^\times(U\cap V)$ such that $e_V=u_{UV}e_U$ as frames over $U\cap V$. On $U\cap V$ we therefore have \begin{align*} e_U(f_U)=s_L=e_V(f_V)=e_U(u_{UV}f_V) \end{align*} Since $e_U$ is an isomorphism, it follows that $f_U=u_{UV}f_V$ on $U\cap V$. Let $Y\subset X$ be a prime divisor, and let $\eta_Y\in X$ be its generic point. Because $X$ is nonsingular, the local ring $\mathcal O_{X,\eta_Y}$ is a regular local ring. Since $Y$ has codimension one in $X$, this local ring has Krull dimension one. By the standard discrete valuation ring criterion for one-dimensional Noetherian regular local domains, $\mathcal O_{X,\eta_Y}$ is a discrete valuation ring. Let $\mathbb N_0:=\{0,1,2,\dots\}$. We write $\operatorname{ord}_Y:\mathcal O_{X,\eta_Y}\setminus\{0\}\longrightarrow \mathbb N_0$ for its valuation on nonzero regular elements. If $\eta_Y\in U\cap V$, then $u_{UV}$ maps to a unit of $\mathcal O_{X,\eta_Y}$, hence $\operatorname{ord}_Y(f_U)=\operatorname{ord}_Y(u_{UV}f_V)=\operatorname{ord}_Y(f_V)$. Thus the order of vanishing of $s_L$ along $Y$ is independent of the chosen trivialization. [/step]

[step:Define the hyperplane section divisor from the local orders] For each prime divisor $Y\subset X$, choose an open set $U_Y\subset X$ containing the generic point $\eta_Y$ and a local frame of $\mathcal O_X(1)$ on $U_Y$. Let $f_{U_Y}:U_Y\longrightarrow \mathbb A_k^1$ be the corresponding local equation for $s_L$. Since $X$ is irreducible and nonsingular, it is integral, and a nonzero section of an invertible sheaf cannot have zero germ at the generic point of any prime divisor unless it is zero on a nonempty open subset of $X$. Therefore the germ of $f_{U_Y}$ in $\mathcal O_{X,\eta_Y}$ is nonzero. Define $m_Y:=\operatorname{ord}_Y(f_{U_Y})$. By the frame-independence proved above, $m_Y$ depends only on $Y$ and $s_L$, not on $U_Y$ or the chosen frame. The hyperplane section divisor is the divisor $H|_X:=\sum_Y m_Y\,Y$, where the sum ranges over the prime divisors, meaning the irreducible codimension-one closed subvarieties, $Y\subset X$. The hypothesis $\dim X\ge 1$ ensures that this is the relevant divisor language on $X$. Since $s_L$ is nonzero, its zero locus is the proper closed subset $X\cap H$ of the Noetherian variety $X$, so only the codimension-one irreducible components of $X\cap H$ can occur with nonzero coefficient. Hence the displayed sum is finite and defines a divisor on $X$. [/step]

[step:Conclude that all coefficients are nonnegative] For every prime divisor $Y\subset X$, the coefficient of $Y$ in $H|_X$ is $m_Y=\operatorname{ord}_Y(f_{U_Y})$, where $f_{U_Y}$ is a nonzero regular element of the discrete valuation ring $\mathcal O_{X,\eta_Y}$. The valuation of a nonzero regular element in a discrete valuation ring is a nonnegative integer. Therefore $m_Y\in \mathbb N_0$ for every prime divisor $Y\subset X$. Thus every coefficient of $H|_X$ is nonnegative, and so $H|_X$ is an effective divisor on $X$. [/step]