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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 $R[X]$?

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Isn't this more or less by definition? To say that F_n is complementary to the described hypersurface is precisely to say that F_n represents the described functor. – Qiaochu Yuan Jan 12 '10 at 21:11
Just a suggestion: maybe put something like [Cazanave, 0912.2227] in the title? I have no idea who Cazanave is, but I suspect that if I had read this paper I'd have a better chance of remembering the author's name than the arXiv number. – Michael Lugo Jan 12 '10 at 21:29
I think that the statement is true, so it should be more or less by definition. I might have the wrong definitions?! Could you make this a little more explicit, please? – user2146 Jan 13 '10 at 9:25
Michael, I think [Cazanave, 0912.2227] would be too long. If not for this paper/question, then for other papers with more authors. I wanted to make it possible to search the MO for a certain article-id and to find all discussions related to this article. – user2146 Jan 13 '10 at 9:29
up vote 3 down vote accepted

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 $R$.

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Let me denote the resultant just by $f$ and let $R$ be a $k$-algebra. A morphism $\mathrm{spec}(R)\rightarrow F_{n}=\mathrm{spec}(k[...]_{f})$ is the same as a $k$-alg morphism $k[...]\rightarrow R$ such that the image of $f$ is invertible and this morphism gives (and is determined by) the values in $R$ for the variables $a_i,b_i$. I was stupidly fixed to a particular stupid idea and that made me unable to see this elementary statement works. Your post helped me to correct my idea, thanks. – user2146 Jan 14 '10 at 14:14

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