# What is the family derived from the absolute Frobenius on the Hilbert scheme?

Let $f$ be a Hilbert polynomial, and $X := Hilb_h(P^d_{F_p})$ a Hilbert scheme defined over $F_p$. Then there is an absolute Frobenius map $F: X \to X$. I'm even interested in the case $f \equiv 1$, so $X = P^d$.

Which of the following is a sensical question?

1. What is the family over $X$ induced by pulling back the universal family along $F$? Is there a reasonable way to think about it, that makes it clear that it's not just again the universal family?

2. If that family is just the universal family again, then in what sense is the Hilbert scheme universal? (As it seems I've obtained the same family from two different maps, which I thought wasn't supposed to happen.)

-
For $f \equiv 1$ the universal family is the diagonal in $P^d \times P^d$. So, its pullback is the graph of the Frobenius. – Sasha Jun 18 '10 at 11:21
@Sasha: Good example. In terms of the answer below, it corresponds to line bundles which are a $p$th power on the base (equipped with a specified choice of that $p$th root), consistent with its realization inside of $\mathbf{P}^d \times \mathbf{P}^d$ projecting via isomorphism to one factor but not to the other. – BCnrd Jun 18 '10 at 13:51

Indeed since the map on the base is not an isomorphism (apart from silly cases such as when $X$ is empty: the Frobenius endomporphism of a locally finite type $\mathbb{F}_p$-scheme is an isomorphism if and only if the scheme is etale), the pullback cannot be universal. Functorially, $F$ carries a family of closed subschemes of projective space to the family defined (locally over the base) by the equations with coefficients raised to the $p$-power. Concretely, a hypersurface with some coefficient of 1 and another not a $p$-power in the base ring does not occur in the pullback family.