Let \(f \in k[x_0,\ldots,x_n]\) be homogeneous, irreducible, of degree \(d\), and let \(X = V(f) \subseteq \PP^n\). Then \(\deg X = d\).
(Degree of a hypersurface section)
Let \(Y \subseteq \PP^n\) be a projective variety of dimension \(\geq 1\) and \(H\) a hypersurface not containing \(Y\), with \(Y \cap H = Z_1 \cup \cdots \cup Z_s\) the irreducible components. Then \[ \sum_{j=1}^s i(Y, H; Z_j) \deg Z_j = (\deg Y)(\deg H) , \] where \(i(Y, H; Z_j)\) is the intersection multiplicity along \(Z_j\). For distinct plane curves \(Y, Z \subseteq \PP^2\) of degrees \(d\) and \(e\) with no common component, \(\sum_{P \in Y \cap Z} i(Y, Z; P) = de\) (Bézout). [@Har10a, Theorem I.7.7]
The proof is the exact sequence \[ 0 \to S(-d) \xrightarrow{\; \cdot f \;} S \to S/(f) \to 0 , \] which is exact because \(S\) is a domain and \(f \neq 0\), so multiplication by \(f\) is injective. Taking dimensions in degree \(r\), \[ P_X(r) = \binom{r+n}{n} - \binom{r-d+n}{n} , \] a polynomial of degree \(n-1\) whose leading coefficient is \(d/(n-1)!\).
The statement is the compatibility of two definitions of degree that look unrelated: the degree of the defining equation, and the normalised leading term of the Hilbert polynomial. The geometric count — the number of points in which \(X\) meets a general line — agrees with both, and that is Bezout in the first nontrivial case.