[proofplan] We restrict the defining homogeneous form $F$ to lines in $\mathbb P_k^n$. The lines on which this restriction is not identically zero form a nonempty Zariski-open subset of the Grassmannian: openness comes from the universal line, and nonemptiness follows by choosing a point where $F$ does not vanish. For any such line $L\cong \mathbb P_k^1$, the scheme-theoretic intersection $X\cap L$ is the divisor on $\mathbb P_k^1$ cut out by a nonzero homogeneous form of degree $m$, hence has length $m$. This length is, by the stated convention, the degree of the projective hypersurface. [/proofplan]

[step:Find a nonempty open set of lines on which $F$ does not vanish identically]Let \begin{align*} G:=\operatorname{Gr}(1,\mathbb P_k^n) \end{align*} be the Grassmannian of projective lines in $\mathbb P_k^n$, and let \begin{align*} \pi:\mathcal L\to G \end{align*} denote the universal line, with evaluation morphism \begin{align*} e:\mathcal L\to \mathbb P_k^n. \end{align*} The [homogeneous polynomial](/page/Homogeneous%20Polynomial) $F$ defines a global section of $\mathcal O_{\mathbb P_k^n}(m)$. Pulling it back along $e$ gives a section \begin{align*} e^*F\in H^0(\mathcal L,e^*\mathcal O_{\mathbb P_k^n}(m)). \end{align*} For each line $L\in G$, the fiber of $\pi$ over $L$ is canonically $L$, and the restriction of $e^*F$ to that fiber is the homogeneous degree-$m$ form $F|_L$ on $L$. Define \begin{align*} U:=\{L\in G:F|_L\ne 0\}. \end{align*} The complement $G\setminus U$ is the locus where the relative section $e^*F$ vanishes identically on the fibers of $\pi$. Since $\pi_*\mathcal O_{\mathcal L}(m)$ is a vector bundle on $G$ and $e^*F$ determines a section of this vector bundle, this vanishing locus is the zero locus of that section. Hence $G\setminus U$ is Zariski closed, so $U$ is Zariski open. It remains to check that $U$ is nonempty. Since $k$ is algebraically closed, it is infinite. A nonzero homogeneous polynomial over an infinite field cannot vanish at every $k$-point of $\mathbb A_k^{n+1}$. Thus there exists a point \begin{align*} a=[a_0:\cdots:a_n]\in \mathbb P_k^n \end{align*} such that $F(a)\ne 0$. If $n=1$, the unique projective line in $\mathbb P_k^1$ is $\mathbb P_k^1$ itself, and $F|_{\mathbb P_k^1}=F\ne 0$, so $U=G$. If $n\ge 2$, choose any projective line $L\subset\mathbb P_k^n$ passing through $a$. Since $F(a)\ne 0$, the restriction $F|_L$ cannot be the zero section on $L$. Therefore $L\in U$, and $U$ is nonempty.[/step]

[guided]The purpose of this step is to isolate the lines that are not contained in the hypersurface. A line $L$ is bad precisely when the restriction $F|_L$ is the zero polynomial, because then every point of $L$ lies in the vanishing locus of $F$ scheme-theoretically. Let \begin{align*} G:=\operatorname{Gr}(1,\mathbb P_k^n) \end{align*} be the Grassmannian whose points are projective lines $L\subset \mathbb P_k^n$. Let \begin{align*} \pi:\mathcal L\to G \end{align*} be the universal line. This means that the fiber $\pi^{-1}(L)$ over a point $L\in G$ is exactly the projective line $L$. There is also an evaluation morphism \begin{align*} e:\mathcal L\to \mathbb P_k^n \end{align*} which sends a pair consisting of a line and a point on that line to the corresponding point of $\mathbb P_k^n$. The polynomial $F$ is homogeneous of degree $m$, so it defines a global section of $\mathcal O_{\mathbb P_k^n}(m)$. Pulling this section back along $e$ gives \begin{align*} e^*F\in H^0(\mathcal L,e^*\mathcal O_{\mathbb P_k^n}(m)). \end{align*} On the fiber over a line $L$, this section is exactly the restricted homogeneous form $F|_L$. Now define \begin{align*} U:=\{L\in G:F|_L\ne 0\}. \end{align*} We show that $U$ is open. The relative sheaf $\pi_*\mathcal O_{\mathcal L}(m)$ is a vector bundle on $G$: its fiber over $L$ is the finite-dimensional [vector space](/page/Vector%20Space) $H^0(L,\mathcal O_L(m))$. The section $e^*F$ determines a section of this vector bundle, whose value at $L$ is the element $F|_L\in H^0(L,\mathcal O_L(m))$. Therefore the bad locus \begin{align*} G\setminus U=\{L\in G:F|_L=0\} \end{align*} is the zero locus of a vector-bundle section. Zero loci of sections of vector bundles are Zariski closed, so $U$ is Zariski open. We also need $U$ to be nonempty. Since $k$ is algebraically closed, it is infinite. A nonzero polynomial over an infinite field cannot vanish as a function on all [affine space](/page/Affine%20Space). Hence the nonzero homogeneous form $F$ has some nonzero vector $(a_0,\dots,a_n)\in k^{n+1}$ with \begin{align*} F(a_0,\dots,a_n)\ne 0. \end{align*} This vector determines a point \begin{align*} a=[a_0:\cdots:a_n]\in \mathbb P_k^n \end{align*} with $F(a)\ne 0$. If $n=1$, the Grassmannian of projective lines in $\mathbb P_k^1$ has only one point, corresponding to $\mathbb P_k^1$ itself. Since $F$ is nonzero, this unique line belongs to $U$. If $n\ge 2$, choose a projective line $L\subset\mathbb P_k^n$ through $a$. Such a line exists because projective space has dimension at least $2$. Since $F(a)\ne 0$, the restriction $F|_L$ cannot be the zero section: a zero section would vanish at every point of $L$, including $a$. Thus $L\in U$. This proves that $U$ is nonempty.[/guided]

[step:Identify the scheme-theoretic intersection with a divisor on $\mathbb P_k^1$] Let $L\in U$. Choose an isomorphism of projective schemes \begin{align*} \varphi:\mathbb P_k^1\to L. \end{align*} Let $s,t$ be homogeneous coordinates on $\mathbb P_k^1$. The pullback of $F|_L$ along $\varphi$ is a nonzero homogeneous polynomial \begin{align*} G_L:=\varphi^*(F|_L)\in k[s,t] \end{align*} of degree $m$. The scheme-theoretic intersection $X\cap L$ is obtained by pulling the defining ideal $(F)$ of $X$ back to $L$. Under the chosen isomorphism $\varphi$, this gives \begin{align*} X\cap L\cong \operatorname{Proj}(k[s,t]/(G_L)). \end{align*} Since $G_L\ne 0$, this closed subscheme of $\mathbb P_k^1$ is a proper effective Cartier divisor, hence is zero-dimensional. [/step]

[step:Compute the length of a nonzero degree-$m$ divisor on $\mathbb P_k^1$] We prove the standard length computation needed in this setting. Let $G\in k[s,t]$ be a nonzero homogeneous polynomial of degree $m$. Since $k$ is algebraically closed, $G$ factors as \begin{align*} G=c\ell_1^{e_1}\cdots \ell_r^{e_r}, \end{align*} where $c\in k^\times$, where $\ell_1,\dots,\ell_r\in k[s,t]$ are pairwise non-proportional nonzero linear forms, where $e_1,\dots,e_r\in \mathbb N$, and where \begin{align*} e_1+\cdots+e_r=m. \end{align*} For each $1\le i\le r$, let \begin{align*} q_i:=V_+(\ell_i)\in \mathbb P_k^1 \end{align*} be the point cut out by $\ell_i$. Let $D:=\operatorname{Proj}(k[s,t]/(G))$. The support of $D$ is the finite set $\{q_1,\dots,q_r\}$. Fix $i$. In the local ring $\mathcal O_{\mathbb P_k^1,q_i}$, the form $\ell_i$ is a local parameter up to multiplication by a unit, while each $\ell_j$ with $j\ne i$ is a unit because $\ell_j(q_i)\ne 0$. Therefore \begin{align*} \mathcal O_{D,q_i}\cong \mathcal O_{\mathbb P_k^1,q_i}/(\ell_i^{e_i}). \end{align*} Choosing an affine coordinate $u_i$ at $q_i$ for which $\ell_i$ is a unit multiple of $u_i$, this local Artin ring is isomorphic to \begin{align*} k[u_i]_{(u_i)}/(u_i^{e_i}). \end{align*} Its $k$-basis is represented by \begin{align*} 1,u_i,\dots,u_i^{e_i-1}, \end{align*} so \begin{align*} \operatorname{length}_{\mathcal O_{\mathbb P_k^1,q_i}}(\mathcal O_{D,q_i})=e_i. \end{align*} Adding the local lengths over all points of the zero-dimensional scheme $D$ gives \begin{align*} \operatorname{length}(D)=\sum_{i=1}^r e_i=m. \end{align*} Applying this to $G=G_L$ gives \begin{align*} \operatorname{length}(X\cap L)=m \end{align*} for every $L\in U$. [/step]

[step:Conclude the degree computation from the general line section] For every line $L$ in the nonempty Zariski-open subset $U\subset \operatorname{Gr}(1,\mathbb P_k^n)$ constructed above, the scheme-theoretic intersection $X\cap L$ is zero-dimensional and has length $m$. Since $X\subset\mathbb P_k^n$ is a hypersurface, its complementary linear sections are projective lines. By the convention that the degree of a projective scheme is computed as the length of a general complementary linear section, we obtain \begin{align*} \deg X=m. \end{align*} This proves the theorem. [/step]