most recent 30 from http://mathoverflow.net 2013-05-22T05:15:26Z http://mathoverflow.net/feeds/question/36208 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36208/diagonalization-of-quadratic-forms-over-euclidean-rings K.J. Moi 2010-08-20T18:03:07Z 2010-08-20T18:41:36Z

<p>Let $A$ be a commutative euclidean ring, (probably) with 2 a unit in $A$. I'm trying to compute Witt and Grothendieck-Witt rings and since $A$ is a PID any f.g.p. module over it is free so I only need think about forms on $A^n$. </p> <p>Question: Is every (non-degenerate) quadratic form over $A$ diagonalizable?</p> <p>A form $q$ is diagonalizable we can perform a base change on $A^n$ such that the matrix for $q$ becomes diagonal. Edit: Non-degenerate here means that any matrix associated to the form is invertible.</p> <p>From Milnor-Husemoller's book I know this is true if $A$ is local. If I can show that any non-degenerate quadratic form on $A^n$ represents some unit ($q(x) = u$ a unit for some $x \in A^n$) then the statement holds by cor. I.3.3 in Mil-Hus. </p>

http://mathoverflow.net/questions/36208/diagonalization-of-quadratic-forms-over-euclidean-rings/36210#36210 Robin Chapman 2010-08-20T18:29:28Z 2010-08-20T18:29:28Z

<p>Quadratic forms over $\mathbb{Z}$ don't diagonalize in general. Even positive definite rank two forms like $3x^2+2xy+5y^2$ can't be diagonalized. Inverting $2$ won't help things.</p>