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 $A^n$. 


Question: Is every (non-degenerate) quadratic form over $A$ diagonalizable?


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.


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. 



Diagonalization of quadratic forms over euclidean rings
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.


