A rational point in the scheme of pointed degree n rational functions [0912.2227]

Question

The following question is related to "Remark 2.2" in Christophe Cazanave's paper "Algebraic homotopy classes of algebraic functions". I decided to add the arxiv article-id to the questions title to invite other people who like to study this article to do the same. My hope is that this will lead to a culture of discussing arxiv articles on the overflow.

Question: Let $F_n$ be the open subscheme of $\mathbb{A}^{2n}=\mathrm{Spec}(k[a_{0},\ldots,a_{n-1},b_{0},\ldots,b_{n-1}])$ complementary to the hypersurface of equation $res_{n,n}(X^{n}+a_{n-1}X^{n-1}+\ldots+a_{0},b_{n-1}X^{n-1}+\ldots+b_{0})$. Let $R$ be a ring. The claim is that an $R$-point of $F_{n}$ is a pair $(A,B)$ of polynomials of $R[X]$, where $A$ is monic of degree $n$, $B$ is of degree strictly less than $n$ and the scalar $res_{n,n}(A,B)$ is invertible. How can I see that a morphism $\mathrm{Spec}(R)\rightarrow F_n$ gives (and is the same as) a pair of polynomials in $R[X]$?

Answer 1

I'm not so good on the scheme-theoretic language, so let me embed $F_n$ as the affine variety $\text{res}\_{n,n}(X^n + ..., b_{n-1} X^{n-1} + ...) y = 1$ one dimension up. Then a morphism $k[a_0, ... a_{n-1}, b_0, ... b_{n-1}, y]/(\text{stuff}) \to R$ is precisely (assuming that Cazanava means either $k = \mathbb{Z}$ or $R$ a $k$-algebra) a choice, for each variable $a_i, b_i, y$, of an element of $R$ subject to the condition that the resultant times $y$ is equal to $1$, i.e. the resultant is invertible in $R$.