I'm curious about the following: What are some arithmetic application of the Hilbert Schemes?

The application of Hilbert schemes in algebraic geometry seems to be a great success, from birational geometry to enumerative geometry. However, although the Hilbert scheme is defined over $\mathbb{Z}$, I haven't seen any use of it in the arithmetic side. If you know such an example please let me know.

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    $\begingroup$ Dear 36min, Picard schemes (for example) are usually constructed using Hilbert schemes, and it is very common in arithmetic to consider Picard schemes of a curve over an integer ring, or over a DVR. Regards, Matthew $\endgroup$
    – Emerton
    Oct 28, 2012 at 3:49
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    $\begingroup$ As a reference, if you look in Chapter 6 of Mumford's GIT, you'll see him rather directly using Hilbert schemes to build moduli schemes (over localized integer rings) of polarized abelian schemes (which in turn underlie PEL Shimura varieties). This is a specific instance of what Liedtke alludes to early in his answer. In these and other applications, it is crucial that fixing the Hilbert polynomial defines a moduli scheme that is finite type (and not merely locally of finite type) over the base. This relative quasi-compactness is an important output of the construction of Hilbert schemes. $\endgroup$
    – user27056
    Oct 28, 2012 at 8:27
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    $\begingroup$ A specific application of moduli schemes (or stacks) might be Faltings's theorem. On the other hand, if you're so inclined I think it's rather a good idea to think about direct applications of Hilbert schemes to arithmetic. They are, after all, more elementary than moduli schemes. There are still many important difficult questions about the arithmetic of moduli schemes, for example, effective versions of Shafarevich's conjecture. It's conceivable that a careful study of the arithmetic of Hilbert schemes could be helpful with them. $\endgroup$ Oct 28, 2012 at 15:24

1 Answer 1


Quite generally, whenever you ``need a moduli space'', say, polarized deformations of varieties, or spaces of morphisms, you oftentimes construct it as follows: first, you construct some family in projective space that (over-)parametrizes your data. Then, using that you've fixed some numerical invariants, you prove that the Hilbert polynomial of this projective family is constant. Then, you use the Hilbert scheme to realize this family as a subscheme of a Hilbert scheme. This is a first approximation to your moduli problem. Usually, you've overparametrized your data and need to take some appropriate stack/GIT quotient (which is usually subtle)...

Thus, in case you're interested in arithmetic moduli, say, moduli spaces of polarized Abelian varieties over $\mathbb{Z}$, moduli spaces of curves of genus $g\geq2$ over $\mathbb{Z}$, you will need the fact that the Hilbert scheme is defined over $\mathbb{Z}$.

Let me even give an application to complex geometry: when proving the existence of rational curves on (complex!) varieties, whose $K_X$ is not nef, via "bend and break", you do the following: you reduce your variety modulo positive characteristic $p$, and construct the desired rational curves on infinitely many reductions modulo $p$ using the Frobenius morphism and characteristic-$p$-methods. Then, you bound the degree of these rational curves (w.r.t. some polarization). Now, to conclude the existence of a rational curve in characteristic zero, you use the space of morphisms from $\mathbb{P}^1$ to show lifting of these curves from characteristic $p$ to characteristic zero. Here, it is essential that this space of morphisms (whose existence relies on the Hilbert scheme) is defined over some ring of integers.

  • $\begingroup$ Very good example! Maybe one should add: we need boundedness of degree so that this "space of morphisms" is of finite type. $\endgroup$ Oct 28, 2012 at 17:20

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