Let us fix some notations:
- $X = \{ F = 0 \} \subset \mathbb{P}^n$ is a non singular hypersurface of degree $d$
- $K_X$ is the canonical bundle of $X$
- $Q(X, K_X) = \{f/g \; \text{such that} \; f,g \in \Gamma(X, K_X^{\otimes p}) \; \text{for some $p > 0$}, g \neq 0 \}$
- $K(X) = deg tr_{\mathbb{C}}Q(X,K_X)$ is the Kodaira dimension of $X$
What I want to show is that $$K(X) = \left\{ \begin{array}{lr} -\infty & \text{if $d < n+1$}\\ 0 & \text{if $d = n+1$}\\ dim_{\mathbb{C}}(X) & \text{if $d > n+1$} \end{array} \right. $$
From adjunction formula we know that $K_X = \mathcal{O}_{\mathbb{P}^n}(d-n-1) \rvert_X$, so the first two cases are easy: if $d < n+1$ we have that $\Gamma(X, K_X^{\otimes p}) = 0$ for all $p >0$; if $d = n+1$ we have that $\Gamma(X, K_X^{\otimes p}) = \mathbb{C}$ for all $p >0$.
I am now dealing with the last case. I know that $\Gamma(X, K_X^{\otimes p}) =$ {Homogeneous polynomial of degree $p(d-n-1)$ restricted to $X$}, so
Initial post $$Q(X,K_X) = \{ f/g \; \text{with $f,g \in \mathbb{C}[z_0, \dots, z_{n-1}]$ of the same degree} \}$$
(I deleted the n-th coordinate because without loss of generality we can suppose that $\partial F / \partial z_n \neq 0$)
At this point I notice that $Q(X,K_X) = Q(\mathbb{P}^{n-1}, \mathcal{O}_{\mathbb{P}^{n-1}}(1))$ and from this I obtain $$K(X) = deg tr_{\mathbb{C}}(Q(\mathbb{P}^{n-1}, \mathcal{O}_{\mathbb{P}^{n-1}}(1)) \leq n-1 = dim_{\mathbb{C}}(X)$$
Now I am searching for $n-1$ elements in $Q(X,K_X)$ algebraically independent. I guess the elements $\{z_i/z_0 \; i=1, \dots, n-1\}$ should work. My attempt to show they are algebraically independent: let $p \in \mathbb{C}[X_1, \dots, X_{n-1}]$ be such that $p(z_1/z_0, \dots, z_{n-1}/z_0) \equiv 0$ and let $q$ be the maximum exponent of the monomials in $p$. Then $z_0^q \cdot p(z_1/z_0, \dots z_{n-1}/z_0)$ is a homogeneous polynomial of degree $q$ in the variables $z_0, \dots, z_{n-1}$ and it's identically zero. Then by the polynomial identity principle its coefficients must be all zero, consequently $p \equiv 0$. Is it correct? (In truth I am only changing name to the variables, so I think it should works. Does it?)
Edit $$Q(X,K_X) = \{ f/g \; \text{with $f,g$ homogeneous polynomials of the same degree}\}$$
I don't ask $f,g$ to be of degree $p(d-n-1)$ for some $p>0$ because if $f,g$ are any two homogeneous polynomials of the same degree $s$ then I can consider
$$\frac{f}{g} = \frac{z_0^{p(d-n-1)-s} f}{z_0^{p(d-n-1)-s} g}$$
Where $p(d-n-1)-s >0$.
With a friend of mine we tried to conclude the proof but we didn't make it right. Here's what we have done so far:
Let $z_i$ be a coordinate in $\mathbb{P}^n$ such that $z_i$ is not identically zero on $X$. Without loss of generality we can suppose $i=0$. Let $z_j$ be a coordinate that is in $F$. Without loss of generality we can suppose $j=1$. Consider the functions $z_i/z_0$ for $i>1$ and suppose it exists a polynomials in $\mathbb{C}[t_1, \dots, t_{n-1}]$ such that $p(z_2/z_0, \dots, z_n/z_0) = 0$. Let $s$ be the degree of $p$, then $q(z_0, \dots, z_n) = z_0^s p(z_2/z_0, \dots, z_n/z_0)$ is a homogeneous polynomial of degree $s$ that vanishes on $X$. Being $F$ non singular and homogeneous it is irreducible, so $F \mid q$ but this is impossible because $F$ depends on $z_j$ and $q$ does not.
Does this proof work? If not, can someone give me a hint? Thank you in advance!