[proofplan] The proof computes the canonical bundle of $X_d$ by applying the [adjunction formula](/theorems/3878) to the smooth hypersurface inclusion $X_d \hookrightarrow \mathbb{P}^n_k$. Since a degree $d$ hypersurface has normal bundle $\mathcal{O}_{X_d}(d)$ and $K_{\mathbb{P}^n_k} \simeq \mathcal{O}_{\mathbb{P}^n_k}(-n-1)$, adjunction gives $K_{X_d} \simeq \mathcal{O}_{X_d}(d-n-1)$. The three cases then follow from the ampleness of positive tensor powers of the restricted hyperplane bundle. [/proofplan] [step:Compute the normal bundle of the degree $d$ hypersurface] Let $i: X_d \hookrightarrow \mathbb{P}^n_k$ denote the closed immersion. Since $X_d$ is a smooth hypersurface of degree $d$, it is an effective Cartier divisor cut out by one homogeneous polynomial of degree $d$. Its ideal sheaf in $\mathbb{P}^n_k$ is therefore \begin{align*} \mathcal{I}_{X_d} \simeq \mathcal{O}_{\mathbb{P}^n_k}(-d). \end{align*} Restricting to $X_d$ gives the conormal bundle \begin{align*} \mathcal{I}_{X_d}/\mathcal{I}_{X_d}^2 \simeq \mathcal{O}_{X_d}(-d). \end{align*} Dualizing, the normal bundle $N_{X_d/\mathbb{P}^n_k}$ is \begin{align*} N_{X_d/\mathbb{P}^n_k} := \mathcal{H}om_{\mathcal{O}_{X_d}}(\mathcal{I}_{X_d}/\mathcal{I}_{X_d}^2,\mathcal{O}_{X_d}) \simeq \mathcal{O}_{X_d}(d). \end{align*} [guided] Let $i: X_d \hookrightarrow \mathbb{P}^n_k$ be the inclusion map. The point of this step is to identify the line bundle that measures the single normal direction to the hypersurface. Because $X_d$ is a degree $d$ hypersurface, it is cut out by one homogeneous polynomial $F \in k[x_0,\dots,x_n]$ of degree $d$. The corresponding ideal sheaf is the image of the line bundle $\mathcal{O}_{\mathbb{P}^n_k}(-d)$ under multiplication by $F$, so \begin{align*} \mathcal{I}_{X_d} \simeq \mathcal{O}_{\mathbb{P}^n_k}(-d). \end{align*} Since $X_d$ is smooth, it is in particular an effective Cartier divisor, and its conormal bundle is the restriction of the ideal sheaf modulo its square: \begin{align*} \mathcal{I}_{X_d}/\mathcal{I}_{X_d}^2 \simeq \mathcal{O}_{X_d}(-d). \end{align*} The normal bundle is the dual of the conormal bundle. Therefore \begin{align*} N_{X_d/\mathbb{P}^n_k} := \mathcal{H}om_{\mathcal{O}_{X_d}}(\mathcal{I}_{X_d}/\mathcal{I}_{X_d}^2,\mathcal{O}_{X_d}) \simeq \mathcal{H}om_{\mathcal{O}_{X_d}}(\mathcal{O}_{X_d}(-d),\mathcal{O}_{X_d}) \simeq \mathcal{O}_{X_d}(d). \end{align*} [/guided] [/step] [step:Apply adjunction to compute $K_{X_d}$] By the adjunction formula for a smooth divisor in a smooth variety (citing a result not yet in the wiki: Adjunction Formula), applied to the smooth divisor $X_d \subset \mathbb{P}^n_k$, \begin{align*} K_{X_d} \simeq \left(K_{\mathbb{P}^n_k} \otimes \mathcal{O}_{\mathbb{P}^n_k}(X_d)\right)\big|_{X_d}. \end{align*} The canonical bundle of projective space is \begin{align*} K_{\mathbb{P}^n_k} \simeq \mathcal{O}_{\mathbb{P}^n_k}(-n-1) \end{align*} (citing a result not yet in the wiki: Canonical Bundle of Projective Space), and the line bundle of the divisor $X_d$ is \begin{align*} \mathcal{O}_{\mathbb{P}^n_k}(X_d) \simeq \mathcal{O}_{\mathbb{P}^n_k}(d). \end{align*} Substituting these identifications into adjunction gives \begin{align*} K_{X_d} &\simeq \left(\mathcal{O}_{\mathbb{P}^n_k}(-n-1) \otimes \mathcal{O}_{\mathbb{P}^n_k}(d)\right)\big|_{X_d} \\ &\simeq \mathcal{O}_{\mathbb{P}^n_k}(d-n-1)\big|_{X_d} \\ &\simeq \mathcal{O}_{X_d}(d-n-1). \end{align*} [guided] We now use the standard adjunction formula. For a smooth divisor $X_d$ inside the smooth variety $\mathbb{P}^n_k$, adjunction says \begin{align*} K_{X_d} \simeq \left(K_{\mathbb{P}^n_k} \otimes \mathcal{O}_{\mathbb{P}^n_k}(X_d)\right)\big|_{X_d}. \end{align*} This is the exact place where smoothness is used: it ensures that $X_d$ has a well-defined canonical bundle and that the hypersurface divisor is regular enough for adjunction to apply. The canonical bundle of projective space is \begin{align*} K_{\mathbb{P}^n_k} \simeq \mathcal{O}_{\mathbb{P}^n_k}(-n-1) \end{align*} (citing a result not yet in the wiki: Canonical Bundle of Projective Space). Since $X_d$ is a divisor of degree $d$, its associated line bundle is \begin{align*} \mathcal{O}_{\mathbb{P}^n_k}(X_d) \simeq \mathcal{O}_{\mathbb{P}^n_k}(d). \end{align*} Putting these two inputs into adjunction yields \begin{align*} K_{X_d} &\simeq \left(K_{\mathbb{P}^n_k} \otimes \mathcal{O}_{\mathbb{P}^n_k}(X_d)\right)\big|_{X_d} \\ &\simeq \left(\mathcal{O}_{\mathbb{P}^n_k}(-n-1) \otimes \mathcal{O}_{\mathbb{P}^n_k}(d)\right)\big|_{X_d} \\ &\simeq \mathcal{O}_{\mathbb{P}^n_k}(d-n-1)\big|_{X_d} \\ &\simeq \mathcal{O}_{X_d}(d-n-1). \end{align*} Thus the entire classification reduces to the sign of the integer $d-n-1$. [/guided] [/step] [step:Read off the Fano case from the positivity of $-K_{X_d}$] Assume $d \leq n$. Then $n+1-d \geq 1$. From the canonical bundle computation, \begin{align*} -K_{X_d} \simeq \mathcal{O}_{X_d}(n+1-d). \end{align*} The line bundle $\mathcal{O}_{\mathbb{P}^n_k}(1)$ is ample, and the restriction of an ample line bundle to a closed subvariety is ample (citing a result not yet in the wiki: Restriction of Ample Line Bundles). Hence $\mathcal{O}_{X_d}(1)$ is ample. Since every positive tensor power of an ample line bundle is ample, $\mathcal{O}_{X_d}(n+1-d)$ is ample. Therefore $-K_{X_d}$ is ample, so $X_d$ is Fano. [/step] [step:Read off the Calabi-Yau canonical-bundle case from the zero exponent] Assume $d=n+1$. Then the canonical bundle formula gives \begin{align*} K_{X_d} \simeq \mathcal{O}_{X_d}(d-n-1) = \mathcal{O}_{X_d}(0) \simeq \mathcal{O}_{X_d}. \end{align*} Thus $K_{X_d}$ is isomorphic to the structure sheaf, so $X_d$ is Calabi-Yau in the canonical-bundle sense. [/step] [step:Read off the general type case from the positivity of $K_{X_d}$] Assume $d \geq n+2$. Then $d-n-1 \geq 1$. From the canonical bundle computation, \begin{align*} K_{X_d} \simeq \mathcal{O}_{X_d}(d-n-1). \end{align*} As above, $\mathcal{O}_{X_d}(1)$ is ample, and every positive tensor power of an ample line bundle is ample. Therefore $\mathcal{O}_{X_d}(d-n-1)$ is ample, so $K_{X_d}$ is ample. Since $X_d$ is smooth and projective, the standard characterization of varieties of general type says that $X_d$ is of general type when its canonical bundle is big. An ample line bundle is big, so $K_{X_d}$ is big, and hence $X_d$ is of general type. [/step]