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 $\mathbf{P}^N$...



Hilbert polynomial of $X\times P^1$
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 $\mathbb P^1$.


Computing the Hilbert polynomial of a product of varieties with a product of line bundles is equivalent to computing the Euler characteristic of that product, which by the Kunneth formula is the product of the Euler characteristics of the two original varieties.


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 $dx+1$.


