The following question is related to "Remark 2.2" in Christophe Cazanave's paper "[Algebraic homotopy classes of algebraic functions](http://arxiv.org/abs/0912.2227v1)". 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 $R[X]$?



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


