# Hilbert polynomial of $X\times P^1$

Let $X$ be a canonically polarized smooth projective geometrically connected variety over $k$ with Hilbert polynomial $h$.

What is the Hilbert polynomial of $X\times_k \mathbf{P}^1_k$? How does it depend on $h$?

Example. Let $g\geq 2$ be an integer and let $X$ be a genus $g$ curve. Then the Hilbert polynomial of $X\times_k \mathbf{P}^1_k$ should depend only on $g$.

Probably one should use some "canonical" embedding of $\mathbf{P}^n \times\mathbf{P}^1$ into $\mathbf{P}^N$...

• A Hilbert polynomial depends on a choice of ample line bundle... – Daniel Litt Dec 21 '12 at 20:24
• Yes. I thought that when one says "canonically polarized" this means we choose some power of the canonical divisor to define the Hilbert polynomial, no? – Mike Lowrey Dec 21 '12 at 20:26
• Yes--but the canonical divisor on $X\times \mathbb{P}^1$ might not be ample. – Daniel Litt Dec 21 '12 at 20:37
• Is the canonical divisor of $X \times \mathbb P^1$ ever ample? – Will Sawin Dec 21 '12 at 20:40
• Isn't $K_{X\times \mathbf{P}^1}|_{X\times a} = K_X\oplus K_{\mathbf{P}^1,a}$ for all $a$ in $\mathbf{P}^1$. Thus, this would show that the canonical sheaf is never ample because it contains a direct factor. – Mike Lowrey Dec 21 '12 at 20:46

Rather than using a canonical bundle, the obvious choice is the Segre embedding $\mathbb P^n \times \mathbb P^1 \to \mathbb P^{2n+1}$. This is equivalent to choosing the ample line bundle that is tensor product of the pullback of the original ample line bundle with the pullback of the line bundle of degree $1$ on $\mathbb P^1$.
Thus, the Hilbert polynomial of the product is the product of the Hilbert polynomials. So, for the Segre embedding, this is just the Hilert polynomial of $X$ times $x+1$. If you chose a degree $d$ divisor on $\mathbb P^1$, it would be the Hilbert polynomial of $X$ times $dx+1$.