most recent 30 from http://mathoverflow.net 2013-05-19T00:25:15Z http://mathoverflow.net/feeds/question/11583 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/11583/a-rational-point-in-the-scheme-of-pointed-degree-n-rational-functions-0912-2227 philip314 2010-01-12T21:06:14Z 2010-01-13T14:32:07Z

<p>The following question is related to "Remark 2.2" in Christophe Cazanave's paper "<a href="http://arxiv.org/abs/0912.2227v1" rel="nofollow">Algebraic homotopy classes of algebraic functions</a>". 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.</p> <p><strong>Question:</strong> 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 $R[X]$?</p>

http://mathoverflow.net/questions/11583/a-rational-point-in-the-scheme-of-pointed-degree-n-rational-functions-0912-2227/11656#11656 Qiaochu Yuan 2010-01-13T13:39:51Z 2010-01-13T14:32:07Z

<p>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 $R$. </p>