# Kodaira dimension of $X \subset \mathbb{P}^n$ hypersurface of degree $d$

Let us fix some notations:

1. $X = \{ F = 0 \} \subset \mathbb{P}^n$ is a non singular hypersurface of degree $d$
2. $K_X$ is the canonical bundle of $X$
3. $Q(X, K_X) = \{f/g \; \text{such that} \; f,g \in \Gamma(X, K_X^{\otimes p}) \; \text{for some$p > 0$}, g \neq 0 \}$
4. $K(X) = deg tr_{\mathbb{C}}Q(X,K_X)$ 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!

• I wouldn't expect this to be true, given that (1) Kodaira dimension is a birational invariant, (2) every variety is birational to a hypersurface in projective space, and (3) there exist varieties of all possible Kodaira dimension. – Tabes Bridges Sep 17 '16 at 16:38
• I can't get the first two points you mentioned because I have not yet studied birational invariants. About the third I don't get the point: in this way you can obtain every dimension simply by taking an hypersurface oF degree greater than n+1 in $\mathbb {P}^n$ with $n$ the desired dimension. However where my attempt should then be wrong? Thank you for the answer! – Federico Sep 17 '16 at 16:46
• Tabes' comment refer, for example, to the fact that an elliptic surface $S$ has kodaira dimension $1$. Now, point (2) in his comment says that there exists an hypersurface in $\mathbb{P}^3$ birational to $S$ and point (1) tells you that the kodaira dimension of this hypersurface is $1$, which contradicts your claim. – User3773 Sep 17 '16 at 17:46
• The first error you are making is in your description of $Q(X,K_X)$. Merely because the partial derivative is non-zero, you can not eliminate $z_n$. Secondly, you are not looking at $f/g$ with $f,g$ of same degree, but $f,g$ are of degree $p(d-n-1)$ for some $p$. – Mohan Sep 17 '16 at 18:31
• Mmm in the statement "any variety is birational to an hypersurface" the latter can be singular. I think you are right, or better I have not been into the third case, whose claim is surely right, and the first two cases seem right as well. – User3773 Sep 17 '16 at 18:32