Let $R \to A$ be a homomorphism of commutative rings. Define the functor $$\mathrm{Hilb}^n_{A/R} : \mathsf{CAlg}(R) \to \mathsf{Set}$$ as follows: If $B$ is a commutative $R$-algebra, then $\mathrm{Hilb}^n_{A/R}(B)$ is the set of surjective $B$-algebra homomorphisms $B \otimes_R A \to Q$, where the underlying $B$-module of $Q$ is locally free of rank $n$.

In the paper "An elementary, explicit, proof of the existence of Hilbert schemes of points" (arXiv, published) it is shown that $\mathrm{Hilb}^n_{A/R}$ is represented by a scheme. But I don't understand how to construct the open covering (sections 5.1 and 5.2):

If $F$ is a free $R$-module of finite rank and $e \in F$ is an element which is part of a basis, and $\beta : F \to A$ is an $R$-linear map such that $\beta(e)=1$, then the authors define $\mathrm{Hilb}^{\beta}_{A/R}$ to be the subfunctor of $\mathrm{Hilb}^n_{A/R}$ which consists of those $B \otimes_R A \to Q$ such that the composition $B \otimes_R F \to B \otimes_R A \to Q$ is surjective.

Q1. The authors claim that the composition has to be an isomorphism. This would be only clear to me if $F$ was free of rank $n$. But I cannot find this assumption anywhere.

Q2. Why is $\mathrm{Hilb}^{\beta}_{A/R}$ an open subfunctor of $\mathrm{Hilb}^n_{A/R}$?

Q3. Why do the functors $\mathrm{Hilb}^{\beta}_{A/R}$ cover $\mathrm{Hilb}^n_{A/R}$?

Although section 5.2 should answer Q2 and Q3, it seems that the authors only reformulate what has to be proven. Also, it is quite confusing that they write for Q3 "we can even choose the $\beta$ to map a fixed basis of $F$ to a given set of generators of $R$". I've also tried to answer Q2 and Q3 for myself, but didn't succeed.

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    $\begingroup$ I recommend that you e-mail Roy Skjelnes. I bet he would be happy to amplify what is in the article. $\endgroup$ Jul 26, 2014 at 18:56
  • $\begingroup$ In my personal opinion, I think it's bad practice to try and understand a potential gap in someone else's paper by asking a question in a public forum. What would happen if a gap or mistake is indeed found? It would only be fair for the authors to have a little time to think about it before being publicly "exposed". $\endgroup$ Sep 28, 2014 at 2:28
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    $\begingroup$ On the other hand, I don't want to disturb authors will silly questions. $\endgroup$ Sep 28, 2014 at 11:10

1 Answer 1


I can now answer the questions: Q1: $F$ has to be free of rank $n$ in the whole paper. Q2: One uses that a homomorphism between locally free modules of rank $n$ is an isomorphism iff its determinant is a unit. Q3: Here $B$ is a field. First one shrinks the generating set $\{\phi(1 \otimes a) : a \in A\}$ to some $B$-basis of $Q$. Then one uses Steinitz exchange lemma in order to get a basis which includes $\phi(1 \otimes 1)$.


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