The canonical bundle of projective space and hypersurfaces

Let $\P^n$ be projective $n$-space with coordinates $[x_0:\cdots:x_n]$. Cover $\P^n$ with affine pieces $U_i = \{x_i\neq 0\}$, each of which are $\A^n$, in coordinates $(y_1,\dots,y_n)$, where $y_j = x_j/x_i$. Recall that the canonical bundle of $\P^n$ is the $n$-fold wedge of the cotangent bundle of $\P^n$, or $\omega_{\P^n} = \bigwedge^nT^*_{\P^n}$. The canonical bundle for an arbitrary variety is defined analogously.

Definition: Let $X$ be a projective $n$-dimensional variety. The sheaf of regular functions on $X$ is $\mathcal O_X$, with $\mathcal O_X(U)=\{f/g\ :\ f,g\in k[x_1,\dots,x_n]/I(X), g\neq 0\}$, and the restriction maps are function restriction.

There is a natural grading on $\mathcal O_X$, given by $\deg(f)-\deg(g)$. A shift in the grading may be applied, called a {\it Serre twist}, to get a differently graded (but isomorphic) module: for $\varphi\in \mathcal O_X$ with $\deg(\varphi)=k$, set $\varphi\in\mathcal O_X(\ell)$ to have $\deg(\varphi) = k-\ell$.

Let $\alpha = dy_1\wedge\cdots\wedge dy_n\in \omega_{\P^n}$, which is well-defined on all of $U_i$. We claim this is well-defined on all of $\P^n$. We check this on the overlap $U_0\cap U_n$ (for nicer notation), but the approach is analogous for $U_i\cap U_j$.
\begin{align*}
U_0 & = \{(y_1,\dots,y_n)\ :\ y_i = x_i/x_0\} & y_i & = \frac{z_{i+1}}{z_i} & dy_i & = \frac{z_1dz_{i+1}-z_{i+1}dz_1}{z_1^2} \\
U_n & = \{(z_1,\dots,z_n)\ :\ z_i = x_{i-1}/x_n\} & y_n & = \frac1{z_1} & dy_n & = \frac{-dz_1}{z_1^2}
\end{align*}
Therefore
\begin{align*}
\alpha & = dy_1\wedge\cdots\wedge dy_n \\
& = \frac{z_1dz_2-z_2dz_1}{z_1^2}\wedge\cdots\wedge \frac{z_1dz_n-z_ndz_1}{z_1^2}\wedge \frac{-dz_1}{z_1^2} \\
& = \frac{dz_2}{z_1}\wedge\cdots\wedge \frac{dz_n}{z_1}\wedge \frac{-dz_1}{z_1^2} \\
& = \frac{(-1)^n}{z_1^{n+1}}dz_1\wedge\cdots \wedge dz_n.
\end{align*}
Since the transition function has a pole of order $n+1$ when $z_1 = 0$, which happens when $x_0=0$, we have that $\alpha$ has a pole of order $n+1$ at $\infty$. Therefore $\omega_{\P^n} \cong \mathcal O_{\P^n}(-n-1)$.

Let $X\subset \P^n$ be a smooth hypersurface defined by a degree $d$ equation $F(x_0,\dots,x_n)=0$. On the affine piece $U_0$ this becomes $f(y_1,\dots,y_n)=F(1,\frac{x_1}{x_0},\dots,\frac{x_n}{x_0})$ with $y_i = x_i/x_0$. The total derivative is
\[
\frac{\dy f}{\dy y_1} dy_1 + \cdots + \frac{\dy f}{\dy y_n} dy_n = \sum_{i=1}^n\frac{\dy f}{\dy y_i}dy_i = 0,
\]
and since $X$ is smooth, the terms never all vanish at the same time. Let $V_i=\{\frac{\dy f}{\dy y_i} \neq 0\}$, and set
\[
\beta_i = \frac{(-1)^{i-1}}{\dy f/\dy y_i} dy_1\wedge\cdots \wedge \widehat{d y_i}\wedge \cdots \wedge d y_n \in \omega_X,
\]
which is well-defined on all of $V_i\subset U_0$. We claim that the choice of $V_i$ does not matter, and indeed, assuming $i<j$,
\begin{align*}
\beta_j & = \frac{(-1)^{j-1}}{\dy f/\dy y_j} dy_1\wedge\cdots \wedge \widehat{d y_j}\wedge \cdots \wedge d y_n \\
& = \frac{(-1)^{j-1+i-1}dy_i}{\dy f/\dy y_j} \wedge dy_1\wedge\cdots \wedge \widehat{d y_i}\wedge \cdots \wedge \widehat{d y_j}\wedge \cdots \wedge d y_n \\
& = \frac{(-1)^{j-1+i-1}\frac{-1}{\dy f/\dy y_i}\left(\frac{\dy f}{\dy y_1}dy_1+\cdots + \widehat{\frac{\dy f}{\dy y_i}dy_i} + \cdots + \frac{\dy f}{\dy y_n}dy_n\right)}{\dy f/\dy y_j} \wedge dy_1\wedge\cdots \wedge \widehat{d y_i}\wedge \cdots \wedge \widehat{d y_j}\wedge \cdots \wedge d y_n \\
& = \frac{(-1)^{j-1+i-1+1}\frac{1}{\dy f/\dy y_i}\cdot \frac{\dy f}{\dy y_j}dy_j}{\dy f/\dy y_j} \wedge dy_1\wedge\cdots \wedge \widehat{d y_i}\wedge \cdots \wedge \widehat{d y_j}\wedge \cdots \wedge d y_n \\
& = \frac{(-1)^{j-1+i-1+1+j-2}}{\dy f/\dy y_i} dy_1\wedge\cdots \wedge \widehat{d y_i}\wedge \cdots \wedge d y_n \\
& = \frac{(-1)^{i-1}}{\dy f/\dy y_i} dy_1\wedge\cdots \wedge \widehat{d y_i}\wedge \cdots \wedge d y_n \\
& = \beta_i.
\end{align*}
Hence $\beta_i$ is well-defined on all of $U_0$, and we call it simply $\beta$. Next we claim it is well-defined on all of $X$. Again we only check on the overlap of $U_0\cap U_n$. On the affine piece $U_n$ this becomes $g(z_1,\dots,z_n)=F(\frac{x_0}{x_n},\dots,\frac{x_{n-1}}{x_n},1)=f(\frac{z_2}{z_1},\dots,\frac{z_n}{z_1},\frac1{z_1})$ with $z_i = x_{i-1}/x_n$. We employ the chain rule $\frac{\dy f}{\dy y_i}=\frac{\dy f}{\dy z_j}\frac{\dy z_j}{\dy y_i}$ and the results above to find that
\begin{align*}
\beta & = \frac{(-1)^{i-1}}{\dy f/\dy y_i} dy_1\wedge\cdots \wedge \widehat{d y_i}\wedge \cdots \wedge d y_n \\
& = \frac{(-1)^{i-1}}{\dy f/\dy z_j \cdot \dy z_j/\dy y_i} \frac{z_1dz_2-z_2dz_1}{z_1^2}\wedge \cdots \wedge \widehat{dy_i}\wedge \cdots \wedge \frac{z_1dz_n-z_ndz_1}{z_1^2}\wedge \frac{-dz_1}{z_1^2} \\
& = \frac{(-1)^{i-1}}{\dy f/\dy z_j \cdot \dy z_j/\dy y_i} \frac{(-1)^{n-1}}{z_1^n}dz_1\wedge\cdots \wedge \widehat{dz_i}\wedge \cdots \wedge dz_n \\
& = \frac{(-1)^{i+n}}{\left(\frac{1}{z_1}\right)^{d-1}\left(c+\cdots\right) z_1^n}dz_1\wedge\cdots \wedge \widehat{dz_i}\wedge \cdots \wedge dz_n \\
& = \frac{(-1)^{i+n}}{z_1^{n-d+1} \left(c+\cdots \right)}dz_1\wedge\cdots \wedge \widehat{dz_i}\wedge \cdots \wedge dz_n,
\end{align*}
for some constant $c$. This comes from expressing $f$ in terms of the $z_i$s and factoring. Since the transition function has a pole of order $n-d+1$ when $z_1 = 0$, which happens when $x_0=0$, we have that $\beta$ has a pole of order $n-d+1$ at $\infty$. Therefore $\omega_{X} \cong \mathcal O_{X}(-n+d-1)$.

References: Griffiths and Harris (Principles of Algebraic Geometry, Chapter 1.2)