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](http://arxiv.org/abs/math/0506161), [published](http://www.sciencedirect.com/science/article/pii/S0022404906002684)) 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.



Open covering of the Hilbert functor of points
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 $\phi(1 \otimes 1)$.


